choosing $3$ non co-planar vectors from set of vector $V = \left\{a \hat\imath+b\hat\jmath+c\hat{k} \mid a,b,c\in \left\{-1,1\right\}\right\}$ Consider the set of $8$ vectors 
$\displaystyle V = \left\{a \hat\imath+b\hat\jmath+c\hat{k}::a,b,c\in \left\{-1,1\right\}\right\}$.
If $3$ non-coplanar vectors are chosen in $2^p$ ways, Then value of $p$ is
My try: Total vectors are
$V=\left\{(1,1,1)\;\;,(1,1,-1)\;\;,(1,-1,1)\;\;,(1,-1,-1)\;\;,(-1,1,1)\;\;,(-1,1,-1)\;\;,(-1,-1,1)\;\;,(-1,-1,-1)\right\}$
Now we can choose $3$ vectors as $\displaystyle \binom{8}{3} = 56$ ways
First we will calculate for co-planar vectors.
If $3$ vectors $\displaystyle a_{i}\hat\imath+b_{i}\hat\jmath+c_{i}\hat{k}\; \forall1\leq  i\leq 3,i\in \mathbb{N}$ are co-planar, Then $\displaystyle \begin{vmatrix}
a_{1} & b_{1} & c_{1}\\ 
a_{2} & b_{2} & c_{2} \\ 
a_{3} & b_{3} & c_{3} 
\end{vmatrix}=0$ 
which is possible when any two row or any two column are zero.
So if we take first two row as $\displaystyle (1,1,1)$, Then  each element of third row as $2$ possibilities .
So total no. of co-planar vectors  are $\displaystyle =2\times 2 \times 2 = 8$
Similarly for second, third rows and same as first, third row 
So we get $8+8+8 = 24$ co-planar vectors.
So non co-planar vectors are $\displaystyle \bf{=\binom{8}{3}-24 = 32 = 2^{5}}$
so we will get $p=5$. 
Is my solution correct?
 A: Any three points in space will be co-planar, so I presume that the exercise means that the origin should not lie in the plane determined by the three chosen points.
Supposing that we want our three points to be $a_1\hat i+b_1\hat j+c_1\hat k,$ $a_2\hat i+b_2\hat j+c_2\hat k,$ and $a_3\hat i+b_3\hat j+c_3\hat k,$ you are correct that we wish to choose these points such that $$\left|\begin{array}{ccc}a_1 & b_1 & c_1\\ a_2 & b_2 & c_2\\ a_3 & b_3 & c_3\end{array}\right|\neq0.$$ However, I am not following your reasoning at all (though your solution is correct). Let me present an alternative approach.
First, observe that if two of the vectors are co-linear with the origin (meaning in this case that they have opposite corresponding components), then the three vectors will be co-planar with the origin, regardless of the other vector. (Why?) There are $4$ ways to pick a pair of vectors from $V$ co-linear with the origin, and $6$ other vectors in $V,$ so there are $4\cdot 6=24$ ways to choose $3$ vectors from $V$ such that $2$ of them are co-linear with the origin. All that remains is to prove that if our three points are co-planar with the origin, then two of them are co-linear with the origin.
To see this, think of the points of $V$ as vertices of a cube centered at the origin, and pick any one of the vertices (call it $v_1$). Note that, except for the vertex co-linear with $v_1$ and the origin, all other vertices lie on a common face with $v_1.$ Pick any of the $3$ faces of which $v_1$ is a corner, and let $v_2$ be any of the other vertices of that face. If $v_1$ and $v_2$ are endpoints of one of the face's $4$ edges, then $v_1$ and $v_2$ are identical in two of their components. Without loss of generality, we may suppose their first two components are equal. (Why?) Now, pick a third vertex ($v_3$), such that our chosen vertices are co-planar with the origin, so that (expanding the determinant by cofactors along the first row) $$\begin{align}0 &= \left|\begin{array}{ccc}a_1 & b_1 & c_1\\ a_2 & b_2 & c_2\\ a_3 & b_3 & c_3\end{array}\right|\\ &= \left|\begin{array}{ccc}a_1 & b_1 & c_1\\ a_1 & b_1 & -c_1\\ a_3 & b_3 & c_3\end{array}\right|\\ &= a_1\bigl(b_1c_3-b_3(-c_1)\bigr)-b_1\bigl(a_1c_3-a_3(-c_1)\bigr)+c_1(a_1b_3-a_3b_1)\\ &= a_1(b_1c_3+b_3c_1)-b_1(a_1c_3+a_3c_1)+c_1(a_1b_3-a_3b_1)\\ &= 2a_1b_3c_1-2a_3b_1c_1\\ &= 2(a_1b_3-a_3b_1)c_1.\end{align}$$ Since $v_1,v_2,v_3$ are distinct elements of $V$ and $v_1,v_2$ match in their first two components, it follows that $a_3=-a_1$ and $b_3=-b_1.$ (Why?) Hence, either $v_3=-v_1$ or $v_3=-v_2,$ as desired. (Why?)
We can use similar reasoning if $v_1$ and $v_2$ are opposite corners of their shared face (in which case they agree in one of their components and are opposite in the other two.)
