$5 x-7 y+z=11$
$6 x-8 y-z=15$
$3 x+2 y-6 z=7$
Given $5 x-7 x+7=11$
$6 x-8 y-z=15$
$3 x+2 y-6 z=7$
$D=\left|\begin{array}{ccc}5 & -7 & 1 \\ 6 & -8 & -1 \\ 3 & 2 & -6\end{array}\right|$
$=5(48+2)+7(-36+3)+1(12+24)$
$=5(50)+7(-33)+1(36)$
$=55$
$D_{1}=\left|\begin{array}{ccc}11 & -7 & 1 \\ 15 & -8 & -1 \\ 7 & 2 & -6\end{array}\right|$
$=11(48+2)+7(-90+7)+1(30+56)$
$=11(50)+7(-83)+1(86)$
$=55$
$D_{2}=\left|\begin{array}{ccc}5 & 11 & 1 \\ 6 & 15 & -1 \\ 3 & 7 & -6\end{array}\right|$
$=5(-90+7)-11(-36+3)+1(42-45)$
$=5(-83)-11(-33)+1(-3)$
$=-55$
$D_{3}=\left|\begin{array}{ccc}5 & -7 & 11 \\ 6 & -8 & 15 \\ 3 & 2 & 7\end{array}\right|$
$=5(-56-30)+7(42-45)+11(12+24)$
$=5(-86)+7(-3)+11(36)$
$=-55$
Now,
$x=\frac{D_{1}}{D}=\frac{55}{55}=1$
$y=\frac{D_{2}}{D}=\frac{-55}{55}=-1$
$z=\frac{D_{3}}{D}=\frac{-55}{55}=-1$
$\therefore x=1, y=-1$ and $z=-1$
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