Given the four vectors in $\mathbb{R}^3$, which combinations are linearly independent? $$\vec{v_1}=\begin{bmatrix}
         2\\-5\\4\\3\\
        \end{bmatrix}
\vec{v_2}=\begin{bmatrix}
         2\\-1\\2\\7\\
        \end{bmatrix}
\vec{v_3}=\begin{bmatrix}
         3\\-1\\4\\5\\
        \end{bmatrix}
\vec{v_4}=\begin{bmatrix}
         3\\-6\\4\\12\\
        \end{bmatrix}
$$
If I construct a matrix $M$ with the 4 given vectors $[\vec{v_1},\vec{v_2},\vec{v_3},\vec{v_4}]$, I can then take the determinant of the matrix.
$$
M = 
\begin{bmatrix}
    2 & 2 & 3 & 3 \\
    -5 & -1 & -1 & -6\\
    4 & 2 & 4 & 4\\
    3 & 7 & 5 & 12
\end{bmatrix}
$$
Since $det(M) = 0$, the given vectors are linearly dependent which implies that at least one of them is a linear combination of the others. I'm not sure how to go about finding that other than to guess and check. 
 A: To find what linear combinations will give a dependency, you can approach via row reduction.
Letting $M=\begin{bmatrix} \mid&\mid&\mid&\mid\\ v_1&v_2&v_3&v_4\\ \mid&\mid&\mid&\mid\end{bmatrix}$, if we row reduce, we can be told which vectors are linearly dependent (building our list from left to right).
In this case, you get $rref(M) = \begin{bmatrix}1&0&0&1\\0&1&0&2\\0&0&1&-1\\0&0&0&0\end{bmatrix}$
This implies that $v_4 = v_1+2v_2-v_3$
The location of the pivots as well implies that the original columns in those locations act as a basis for the span of the vectors (and are thus independent).
In general, if the $k^{th}$ column of the row reduced matrix looks like $\begin{bmatrix}a_1\\a_2\\a_3\\\vdots\end{bmatrix}$ then $v_k = a_1v_1 + a_2v_2 + a_3v_3+\dots$
One could also consider instead $M^T = \begin{bmatrix}- v_1^T -\\
- v_2^T-\\
-v_3^T-\\
-v_4^T-\end{bmatrix}$ and row reduce, it will give us that the resulting nonzero rows (transposed) act as a basis for the span of the set of vectors as well.
