Answer
$x=2$
$y=-3$
$z=4$
Work Step by Step
First we have to compute the determinants $D,D_x,D_y,D_z$:
$D=\begin{vmatrix}4&-5&-6\\1&-2&-5\\2&-1&0\end{vmatrix}=4(0-5)-(-5)(0+10)+(-6)(-1+4)=12$
$D_x=\begin{vmatrix}-1&-5&-6\\-12&-2&-5\\7&-1&0\end{vmatrix}=(-1)(0-5)-(-5)(0+35)+(-6)(12+14)=24$
$D_y=\begin{vmatrix}4&-1&-6\\1&-12&-5\\2&7&0\end{vmatrix}=4(0+35)-(-1)(0+10)+(-6)(7+24)=-36$
$D_z=\begin{vmatrix}4&-5&-1\\1&-2&-12\\2&-1&7\end{vmatrix}=4(-14-12)-(-5)(7+24)+(-1)(-1+4)=48$
We use Cramer's Rule to determine the solutions of the system:
$x=\dfrac{D_x}{D}=\dfrac{24}{12}=2$
$y=\dfrac{D_y}{D}=\dfrac{-36}{12}=-3$
$z=\dfrac{D_z}{D}=\dfrac{48}{12}=4$
The solution is:
$x=2$
$y=-3$
$z=4$
