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If $S$ is the set of all real values of ' $a$ ' such that a plane passing through the points $\left(-\mathrm{a}^2, 1,1\right),\left(1,-\mathrm{a}^2, 1\right)$, $\left(1,1,-\mathrm{a}^2\right)$ also passes through the point $(-1,-1,1)$, then $\mathrm{S}=$
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Verified Answer
The correct answer is:
$\{\sqrt{3},-\sqrt{3}\}$
Since all four points are coplaner
$\begin{aligned}
& \Rightarrow\left|\begin{array}{ccc}
1-a^2 & 2 & 0 \\
2 & -a^2+1 & 0 \\
2 & 2 & -a^2-1
\end{array}\right|=0 \Rightarrow\left(a^2+1\right)^2\left(3-a^2\right)=0 \\
& \Rightarrow a= \pm \sqrt{3}
\end{aligned}$
So, $S=\{-\sqrt{3}, \sqrt{3}\}$
$\begin{aligned}
& \Rightarrow\left|\begin{array}{ccc}
1-a^2 & 2 & 0 \\
2 & -a^2+1 & 0 \\
2 & 2 & -a^2-1
\end{array}\right|=0 \Rightarrow\left(a^2+1\right)^2\left(3-a^2\right)=0 \\
& \Rightarrow a= \pm \sqrt{3}
\end{aligned}$
So, $S=\{-\sqrt{3}, \sqrt{3}\}$
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