Solve this

Given: $3 x-y+2 z=6$

$D=\left|\begin{array}{ccc}3 & -1 & 2 \\ 2 & -1 & 1 \\ 3 & 6 & 5\end{array}\right|$

$3(-5-6)+1(10-3)+2(12+3)=4$

Since $D$ is non-zero, the system of linear equations is consistent and has a unique solution.

$D_{1}=\left|\begin{array}{ccc}6 & -1 & 2 \\ 2 & -1 & 1 \\ 20 & 6 & 5\end{array}\right|$

$=6(-5-6)+1(10-20)+2(12+20)$

$=-66-10+64$

$=-12$

$D_{2}=\left|\begin{array}{ccc}3 & 6 & 2 \\ 2 & 2 & 1 \\ 3 & 20 & 5\end{array}\right|$

$=3(10-20)-6(10-3)+2(40-6)$

$=-30-42+68$

$=-4$

$D_{3}=\left|\begin{array}{ccc}3 & -1 & 6 \\ 2 & -1 & 2 \\ 3 & 6 & 20\end{array}\right|$

$=3(-20-12)+1(40-6)+6(12+3)$

$=-96+34+90$

$=28$

Now,

$x=\frac{D_{1}}{D}=\frac{-12}{4}=-3$

$y=\frac{D_{2}}{D}=\frac{-4}{4}=-1$

$z=\frac{D_{3}}{D}=\frac{28}{4}=7$

$\therefore x=-3, y=-1$ and $\mathrm{z}=7$