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Let the system of linear equations
$x+y+\alpha z=2$
$3 x+y+z=4$
$x+2 z=1$
have a unique solution $\left(x^{*}, y^{*}, z^{*}\right)$. If $\left(\alpha, x^{*}\right),\left(y^{*}, \alpha\right)$ and $\left(x^{*},-y^{*}\right)$ are collinear points, then the sum of absolute values of all possible values of $\alpha$ is
$4$
$3$
$2$
$1$
Solution
$\Delta=\left|\begin{array}{lll}1 & 1 & \alpha \\ 3 & 1 & 1 \\ 1 & 0 & 2\end{array}\right|=-(\alpha+3)$
$\Delta_{1}=\left|\begin{array}{lll}2 & 1 & \alpha \\ 4 & 1 & 1 \\ 1 & 0 & 2\end{array}\right|=-(3+\alpha)$
$\Delta_{2}=\left|\begin{array}{lll}1 & 2 & \alpha \\ 3 & 4 & 1 \\ 1 & 1 & 2\end{array}\right|=-(\alpha+3)$
$\Delta_{3}=\left|\begin{array}{lll}1 & 1 & 2 \\ 3 & 1 & 4 \\ 1 & 0 & 1\end{array}\right|=0$
$\alpha \neq-3, x=1, y=1, z=0$,
Now points $(\alpha, 1),(1, \alpha) \ and (1,-1)$ are collinear
$\left|\begin{array}{lcc}\alpha & 1 & 1 \\ 1 & \alpha & 1 \\ 1 & -1 & 1\end{array}\right|=0$
$\Rightarrow \alpha(\alpha+1)-1(1-1)+1(-1-\alpha)=0$
$\alpha^{2}+\alpha-1-\alpha=0$
