Let the system of linear equations x + y + a z = 2 3 x + y + z = 4 x + 2 z = 1 have a unique solution x , y …

Given,
$x+y+az=2$

$3x+y+z=4$

$x+2z=1$
Now finding,

$\Delta =\left|\begin{array}{llr}1& 1& \alpha \\ 3& 1& 1\\ 1& 0& 2\end{array}\right|=-\left(\alpha +3\right)$

${\Delta }_1=\left|\begin{array}{lll}2& 1& \alpha \\ 4& 1& 1\\ 1& 0& 2\end{array}\right|=-\left(3+\alpha \right)$

${\Delta }_2=\left|\begin{array}{lll}1& 2& \alpha \\ 3& 4& 1\\ 1& 1& 2\end{array}\right|=-\left(\alpha +3\right)$

${\Delta }_3=\left|\begin{array}{lll}1& 1& 2\\ 3& 1& 4\\ 1& 0& 1\end{array}\right|=0$

$\alpha \ne -3,x=\frac{{△}_1}{△}=1,y=\frac{{△}_2}{△}=1,z=\frac{{△}_3}{△}=0,$

Now points $\left(\alpha ,1\right),\left(1,\alpha \right) \& \left(1,-1\right)$ are collinear

$\left|\begin{array}{ccc}\alpha & 1& 1\\ 1& \alpha & 1\\ 1& -1& 1\end{array}\right|=0$

$\Rightarrow \alpha \left(\alpha +1\right)-1\left(1-1\right)+1\left(-1-\alpha \right)=0$

${\alpha }^2+\alpha -1-\alpha =0$

$\alpha =\pm 1$

So, $\left|1\right|+\left|-1\right|=2$