For what values of

We have:

$\left[\begin{array}{lll}1 & 2 & 1\end{array}\right]\left[\begin{array}{lll}1 & 2 & 0 \\ 2 & 0 & 1 \\ 1 & 0 & 2\end{array}\right]\left[\begin{array}{l}0 \\ 2 \\ x\end{array}\right]=0$

$\Rightarrow\left[\begin{array}{lll}1+4+1 & 2+0+0 & 0+2+2\end{array}\right]\left[\begin{array}{l}0 \\ 2 \\ x\end{array}\right]=0$

$\Rightarrow\left[\begin{array}{lll}6 & 2 & 4\end{array}\right]\left[\begin{array}{l}0 \\ 2 \\ x\end{array}\right]=0$

$\Rightarrow[6(0)+2(2)+4(x)]=0$

$\Rightarrow[4+4 x]=[0]$

$\therefore 4+4 x=0$

$\Rightarrow x=-1$

Thus, the required value of x is −1.