A Combinatorial Problem on Families of Subsets of $[n]$ Given a ground set $[n]$, under what condition of parameters $a,b,c$ does a family of subsets $\mathcal{F}\subseteq 2^{[n]}$ with the following property exist?
(i) $\forall S\in \mathcal{F}$, $|S|=a$.
(ii) $\forall S_1,S_2\in \mathcal{F}$, $|S_1\triangle S_2|\ge b$, where $S_1\triangle S_2$ means the symmetric difference between two sets.
(iii) $\forall T \subseteq [n], |T|\ge c$, $\exists S\in \mathcal{F}$, such that $|S\cap T|\ge 0.5a$.
As mentioned by @hardmath in the answer below, the problem is trivial when considering a single block and correspondingly very large $c$. Basically the problem asks for an asympototic solution: Under what condition of $a,b$ we can get $c=o(n)$.
Any known results or related ideas about nontrivial sufficient (no need to be necessary of course) condition of $a,b,c$ is appreciated.
This combinatorial problem rises from a communication game, and may have concrete connection with error-correcting codes.
 A: Normally I try to work with the notation that is proposed in the body of the Question. In this case the request for "known results or related ideas" justifies a comparison of that notation with what is customary in research literature on block designs.  A great resource for constructions and examples of various block designs is the La Jolla Covering Repository.
Conventionally instead of $n$ for the number of elements in the ground set ("points"), this parameter is $v$.  Instead of $a$ for the uniform size of subsets ("blocks"), the "block size" parameter is typically $k$.  [Note: Parameter $b$, used here in property (ii) to restrict the intersections of blocks, is conventionally used for how many blocks there are.]
Property (ii) proscribes the minimum size of the symmetric difference of all pairs of distinct blocks $S_1,S_2$:
$$ |S_1 \bigtriangleup S_2| \ge b $$
Because $S_1 \bigtriangleup S_2 = (S_1 \cup S_2)\setminus (S_1 \cap S_2)$, this is equivalent in the original notation to restricting the maximum size of intersections of any two distinct blocks:
$$ |S_1 \cap S_2| \le \frac{2a-b}{2} $$
Because the blocks are distinct sets of size exactly $a$, the symmetric difference is of even size at most $\min(2a,n)$.  $b=2a$ is achieved only when all the blocks are disjoint.  On the other hand the symmetric difference of two distinct sets of equal size will always contain at least two points.  So every uniform block design satisfies property (ii) for some even $b$ between $2$ and $\min(2a,n)$.
Among the block designs which have been especially well studied, $2$-designs (also called Balanced Incomplete Block Designs) are those for which every pair of distinct points $x,y$ appears in exactly the same fixed number $\lambda \ge 1$ of blocks.  Among BIBDs, those which have a constant size intersection between distinct blocks are called symmetric BIBDs.  For these block designs two distinct blocks always intersect in exactly $\lambda$ points.
Property (iii) asks for a parameter $c$ such that every subset $T \subseteq [n]$ with at least $c$ points has an intersection with some block $S$ containing at least $a/2$ points.  Equivalently one might ask this only about subsets $T$ with exactly $c$ points.  Any block design will satisfy property (iii) for $c$ large enough, e.g. $c=n$.  Therefore it makes sense (as the OP mentions in a Comment on the Question) to find the minimal value of $c$ possible for given values of $n,a,b$.
Observe that any $2$-design (BIBD) of block size $4$ will satisfy property (iii) with $c \ge 2$.  Further a symmetric BIBD with block size $a=4$ will satisfy property (ii) with $b = 2(a - \lambda)$.  An example where $\lambda = 2$ is known, so where $n=7,a = 4,b = 4$ we get minimal $c=2$:
Example: Biplane geometry of order 2
This design uses the complementary blocks of the Fano plane, so here $n=7$ and we take these seven blocks of size $4$:
$$ \{4,5,6,7\},\{1,2,6,7\},\{2,3,4,7\},\{2,3,5.6\},\{1,3,4,6\},\{1,2,4,5\},\{1,3,5,7\} $$
Now block size $a=4$, and each pair of distinct blocks has intersection of size two, so the symmetric differences are all of size $b=4$.  Any pair of distinct points appears in at least one block (actually in two blocks).  So $c = 2 = a/2$ satisfies the property (iii), and this is clearly the least value of $c$ that can do so (since smaller sets cannot have large enough intersections with blocks).
Extremality, Vacuity and Triviality
Most discussions of extremality in block designs touch upon the number of blocks belonging to the designs.  Here that aspect is nearly absent from the definition, although property (iii) does guarantee the inclusion of at least one block.  
However any single-block design satisfies property (ii) vacuously for any value of $b$, so it seems at odds with the spirit of the present investigation.  A design with a single block of size $a \le n$ would technically satisfy all three defining properties when $c \ge n - \lfloor a/2 \rfloor$.  To meet the "nontrivial sufficient" conditions asked about in the Question, the rest of our discussion will assume more than one block is present.
For any $a$-uniform block design on $n$ points there will be values $b,c$ which satisfy properties (ii) and (iii) respectively.  For example $b$ could be the minimum size of all symmetric differences between distinct blocks, and $c=n$ always works.  
Of interest are the largest $b$ and smallest $c$ achieved for a particular design.  To reinforce this, if a given block design satisfies property (ii) for $b$ and property (iii) for $c$, then trivially it will also satisfy (ii) for any $b'\le b$ and (iii) for any $c'\ge c$.
Among all $a$ uniform block designs on $n$ points, however, the largest $b$ and the smallest $c$ may not be attained by the same design.  For example, the above biplane geometry example shows that for $a=4$ and $n=7$, we can have $b=4$ and $c=2$, which is clearly the smallest possible $c$ for such designs.  But the largest $b$ we can achieve is $b=6$, e.g. with a design consisting of two blocks that intersect in a single point:
$$ \{1,2,3,4\},\{4,5,6,7\} $$
This design achieves the largest $b$ (since $n=7$), but at the "loss" of having $c=3$, an easy application of the Pigeonhole Principle.  Clearly we cannot achieve both $b=6$ and $c=2$.
At the other extreme we can achieve $b=2$ and $c= \lceil a/2 \rceil$ by taking as our blocks all the $a$-subsets of $[n]$.  This leads us to ask if designs exist in which $a,b,c$ are of approximately equal magnitude and (as the Question is updated) of $o(n)$.
Existence of block designs with nearly equal a,b,c for n large
We can "constructively" show the existence of block designs where $n$ is arbitrarily large, for any $a \lt n$, in which $b$ is not less than $a$ and $c$ is not greater than $a$.
Consider the $a$ subsets of $[n]$ generated in any sequential order:
$$ S_1, S_2, S_3, \ldots, S_{\binom{n}{a}} $$
Initializing a family of such blocks $\mathcal{F}_0 = \emptyset$, we proceed iteratively to define $\mathcal{F}_{i} = \mathcal{F}_{i-1}$ if there exists $j\lt i$ such that:
$$ |S_i \cap S_j| \ge a/2 $$
and otherwise $\mathcal{F}_{i} = \mathcal{F}_{i-1} \cup \{S_i\}$.  In other words we adjoin the next $a$-subset $S_i$ if and only if it does not meet any of the previous blocks in at least $a/2$ elements.
Let $\mathcal{F} = \mathcal{F}_{\binom{n}{a}}$ be the final result of this procedure.  We now claim these properties:
(i) All blocks in $\mathcal{F}$ are of uniform size $a$.
(ii) The symmetric difference of any two distinct blocks in $\mathcal{F}$ has size at least $a$.
(iii) Any subset $T$ of $[n]$ of size $a$ (or greater) has intersection with some $S \in \mathcal{F}$ of size at least $a/2$.
Claimed property (i) is obvious from our construction of adjoining only $a$-subsets to form $\mathcal{F}$.  
To show (ii), let $S_i,S_j$ be any blocks belonging to $\mathcal{F}$, and without loss of generality $i \gt j$.  Since we added $S_i$ to the family of sets at step $i$, it must be true that $|S_i \cap S_j| \lt a/2$. Thus:
$$ \begin{align*} |S_i \bigtriangleup S_j| &= |S_i \cup S_j| - |S_i \cap S_j| \\
 &= |S_i| + |S_j| - 2|S_i \cap S_j| \\
 &= 2a -2|S_i \cap S_j| \\
 &\gt 2a - 2(a/2) = a \end{align*} $$
To show (iii), let $T$ be any $a$-subset of $[n]$, so that $T = S_i$ for some $1\le i \le \binom{n}{a}$.  If $S_i \in \mathcal{F}$, we are done since $|T \cap S_i| = a \ge a/2$.  If $S_i \not\in \mathcal{F}$, then there exists $j\lt i$ such that $|S_i \cap S_j| \ge a/2$, and again we are done.
We remark that such a block design is quasi-optimal with respect to both maximizing $b$ and minimizing $c$. This is because $b$ is at most $2a$ (and here we achieved $b \gt a$), and $c$ is at least $a/2$ (and here we achieved $c \le a$).
