Solve this

Given: $6 x+y-3 z=5$

$D=\left|\begin{array}{ccc}6 & 1 & -3 \\ 1 & 3 & -2 \\ 2 & 1 & 4\end{array}\right|$

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

$=6(14)-1(8)-3(-5)$

$=91$

$D_{1}=\left|\begin{array}{ccc}5 & 1 & -3 \\ 5 & 3 & -2 \\ 8 & 1 & 4\end{array}\right|$

$=5(12+2)-1(20+16)-3(5-24)$

$=5(14)-1(36)-3(-19)$

$=91$

$D_{2}=\left|\begin{array}{ccc}6 & 5 & -3 \\ 1 & 5 & -2 \\ 2 & 8 & 4\end{array}\right|$

$=6(20+16)-5(4+4)-3(8-10)$

$=6(36)-5(8)-3(-2)$

$=182$

$D_{3}=\left|\begin{array}{lll}6 & 1 & 5 \\ 1 & 3 & 5 \\ 2 & 1 & 8\end{array}\right|$

$=6(24-5)-1(8-10)+5(1-6)$

$=6(19)-1(-2)+5(-5)$

$=91$

Now,

$x=\frac{D_{1}}{D}=\frac{91}{91}=1$

$y=\frac{D_{2}}{D}=\frac{182}{91}=2$

$z=\frac{D_{3}}{D}=\frac{91}{91}=1$

$\therefore x=1, y=2$ and $z=1$