Solution:

$D=\left|\begin{array}{ccc}{1} & {1} & {1} \\ {1} & {-1} & {1} \\ {1} & {1} & {-2}\end{array}\right|=1(2-1)-1(-2-1)+1(1+1)=6$

$D_{x}=\left|\begin{array}{ccc}{3} & {1} & {1} \\ {1} & {-1} & {1} \\ {0} & {1} & {-2}\end{array}\right|=3(2-1)-1(-2-0)+1(1+0)=6$

$\therefore x=\frac{D_{x}}{D}=\frac{6}{6}=1$

$D_{y}=\left|\begin{array}{lll}{1} & {3} & {1} \\ {1} & {1} & {1} \\ {1} & {0} & {-2}\end{array}\right|=1(-2-0)-3(-2-1)+1(0-1)=6$

$\therefore y=\frac{D_{y}}{D}=\frac{6}{6}=1$

$D_{z}=\left|\begin{array}{ccc}{1} & {1} & {3} \\ {1} & {-1} & {1} \\ {1} & {1} & {0}\end{array}\right|=1(0-1)-1(0-1)+3(1+1)=6$

$\therefore z=\frac{D_{z}}{D}=\frac{6}{6}=1$