Finding values of constants when solving linearly dependent equation Suppose we have vectors:-
$X_1=[-2,-2,2] $,  $X_2 = [-2,-5,-1]$, $X_3=[2,-1,-5]$
I found these vectors to be linearly dependent by finding the determinant. The determinant came out to be zero.
If the vectors are linearly dependent, then atleast one of the constants$(k_1,k_2,k_3)$ should be non-zero such that
$k_1X_1 + k_2X_2 +k_3X_3 = 0. $
So i get the following set of equations:-

$-2k_1 -2k_2 +2k_3=0$,
$-2k_1 -5k_2 -k_3=0$,
$ 2k_1 -k_2 -5k_3=0$.

Here I am not able to find even a single non zero constant which makes these set of vectors linearly dependent.

There is a very handy method, namely Gaussian elimination and the reduced row echelon form:
\begin{align}
\begin{bmatrix}
-2 & -2 & 2 \\
-2 & -5 & -1 \\
2 & -1 & -5
\end{bmatrix}
&\to
\begin{bmatrix}
1 & 1 & -1 \\
-2 & -5 & -1 \\
2 & -1 & -5
\end{bmatrix}
&& R_1\gets-\tfrac{1}{2}R_1
\\[6px] &\to
\begin{bmatrix}
1 & 1 & -1 \\
0 & -3 & -3 \\
0 & -3 & -3
\end{bmatrix}
&&\begin{aligned} R_2&\gets R_2+2R_1\\ R_3&\gets R_3-2R_1\end{aligned}
\\[6px] &\to
\begin{bmatrix}
1 & 1 & -1 \\
0 & 1 & 1 \\
0 & -3 & -3
\end{bmatrix}
&& R_2\gets -\tfrac{1}{3}R_2
\\[6px] &\to
\begin{bmatrix}
1 & 1 & -1 \\
0 & 1 & 1 \\
0 & 0 & 0
\end{bmatrix}
&& R_3\gets R_3+3R_2
\\[6px] &\to
\begin{bmatrix}
1 & 0 & -2 \\
0 & 1 & 1 \\
0 & 0 & 0
\end{bmatrix}
&& R_1\gets R_1-R_2
\end{align}
Once you have the reduced row echelon form you can directly read that
$$
x_3=-2x_1+x_2
$$
This happens because row operations don't change the linear relations between the columns; in the last matrix, it is clear that the last column can be written as $-2$ times the first column plus the second column, so the same happens for the original columns.
