Looks like the system is
x + 4y - z = -14
5x + 6y + 3z = 4
-2x + 7y + 2z = -17
or in matrix form,
[tex]\mathbf{Ax} = \mathbf b \iff \begin{bmatrix} 1 & 4 & -1 \\ 5 & 6 & 3 \\ -2 & 7 & 2 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} -14 \\ 4 \\ -17 \end{bmatrix}[/tex]
Cramer's rule says that
[tex]x_i = \dfrac{\det \mathbf A_i}{\det \mathbf A}[/tex]
where [tex]x_i[/tex] is the solution for i-th variable, and [tex]\mathbf A_i[/tex] is a modified version of [tex]\mathbf A[/tex] with its i-th column replaced by [tex]\mathbf b[/tex].
We have 4 determinants to compute. I'll show the work for det(A) using a cofactor expansion along the first row.
[tex]\det \mathbf A = \begin{vmatrix} 1 & 4 & -1 \\ 5 & 6 & 3 \\ -2 & 7 & 2 \end{vmatrix}[/tex]
[tex]\det \mathbf A = \begin{vmatrix} 6 & 3 \\ 7 & 2 \end{vmatrix} - 4 \begin{vmatrix} 5 & 3 \\ -2 & 2 \end{vmatrix} - \begin{vmatrix} 5 & 6 \\ -2 & 7 \end{vmatrix}[/tex]
[tex]\det \mathbf A = ((6\times2)-(3\times7)) - 4((5\times2)-(3\times(-2)) - ((5\times7)-(6\times(-2)))[/tex]
[tex]\det\mathbf A = 12 - 21 - 40 - 24 - 35 - 12 = -120[/tex]
The modified matrices and their determinants are
[tex]\mathbf A_1 = \begin{bmatrix} -14 & 4 & -1 \\ 4 & 6 & 3 \\ -17 & 7 & 2\end{bmatrix} \implies \det\mathbf A_1 = -240[/tex]
[tex]\mathbf A_2 = \begin{bmatrix} 1 & -14 & -1 \\ 5 & 4 & 3 \\ -2 & -17 & 2 \end{bmatrix} \implies \det\mathbf A_2 = 360[/tex]
[tex]\mathbf A_3 = \begin{bmatrix} 1 & 4 & -14 \\ 5 & 6 & 4 \\ -2 & 7 & -17 \end{bmatrix} \implies \det\mathbf A_3 = -480[/tex]
Then by Cramer's rule, the solution to the system is
[tex]x = \dfrac{-240}{-120} \implies \boxed{x = 2}[/tex]
[tex]y = \dfrac{360}{-120} \implies \boxed{y = -3}[/tex]
[tex]z = \dfrac{-480}{-120} \implies \boxed{z = 4}[/tex]
