Find a value r so that the vector v is in the span of a set of vectors Find the value r so that,
$$v = \begin{pmatrix} 3 \\r \\-10\\14 \end{pmatrix}$$
is in the set,
$$ S=  \text{span}\left(\begin{pmatrix} 3\\3\\1\\5
\end{pmatrix}, \begin{pmatrix} 0\\3\\4\\-3 \end{pmatrix}, \begin{pmatrix} 0\\0\\-3\\3 \end{pmatrix} \right) $$
What I've tried is this..., brute force and intuition.  Since we're looking for a linear combination of the 3 vectors in the set S, we know that we can only use a multiple of 1 from the first vector, call this $c_1=1$. We still need $c_2$ and $c_3$ (combinations of the second vector and third vector, respectively).  One can see that if we let $c_2=-2$, and $c_3=1$, we can get the bottom two entries of $v$.  Hence, the only value that $r$ can be is $-3$ , so $r=-3$.   
This is great..., but how about when the system is really complicated and trial and error won't suffice, how would I be able to calculate an arbitrary $r$ that is in a vector $\in$$\mathbb{R}^N$.  With a set of vectors that span $\mathbb{R}^N$ 
 A: Your ad hoc solution is excellent. The general method consists in solving a system of linear equations, reducing it to row echelon form via Gaussian elimination. In your case you start with the matrix
$$
\begin{bmatrix}
0&0&3&3\\
0&3&3&r\\
-3&4&1&-10\\
3&-3&5&14\\
\end{bmatrix}
$$
and apply row elementary operations to get
$$
\begin{bmatrix}
0&0&3&|&3\\
0&3&3&|&r\\
0&1&6&|&4\\
3&-3&5&|&14\\
\end{bmatrix},\quad
\begin{bmatrix}
0&0&3&|&3\\
0&0&-15&|&r-12\\
0&1&6&|&4\\
3&-3&5&|&14\\
\end{bmatrix},\quad
\begin{bmatrix}
0&0&-15&|&r-12\\
0&0&3&|&3\\
0&1&6&|&4\\
3&-3&5&|&14\\
\end{bmatrix},\quad
\begin{bmatrix}
0&0&0&|&r+3\\
0&0&3&|&3\\
0&1&6&|&4\\
3&-3&5&|&14\\
\end{bmatrix}.
$$
This shows that there is a solution only when $r+3 = 0$, that is, $r = -3$.
A: You can use your row reduction skills to solve this. We have 3 equations and 3 unknowns, where the unknowns are $c_1$, $c_2$ and $c_3$.
We have: $3c_1+0c_2+0c_3=3,1c_1+4c_2-3c_3=-10,5c_1-3c_2+3c_3=14$. This suggests the augmented matrix:
$$ \begin{pmatrix} 3&0&0&3\\1&4&-3&-10\\5&-3&3&14 \end{pmatrix} $$
When row reduced, this gives
$$ \begin{pmatrix} 1&0&0&1\\0&1&0&-2\\0&0&1&1 \end{pmatrix} $$
Which means $c_1=1,c_2=-2,c_3=1$.
Then, taking $3c_1+3c_2+0c_3=-3=r$.
This method works in general.
A: You are correct, this problem was set up with the first coordinate nice.  In general, if you have $n$ vectors in $\Bbb R^n$ with one unknown value, you can just form the determinant and choose the value that makes the determinant zero.  This will be a linear equation, but will take a bunch of computation to get there.  In your example, for $n=4$, there are $4!=24$ terms in the determinant.  Six of them will involve $r$ and the other eighteen will not.
