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Let $S$ be the set of all real values of $\lambda$ such that a plane passing through the points $\left(-\lambda^2, 1,1\right),\left(1,-\lambda^2, 1\right)$ and $\left(1,1,-\lambda^2\right)$ also passes through the point $(-1,-1,1)$. Then $S$ is equal to
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The correct answer is:
$\{-\sqrt{3}, \sqrt{3}\}$
from the condition of coplanarity of four points
$\begin{aligned} & \left|\begin{array}{ccc}-\lambda^2+1 & 1+1 & 1-1 \\ 1+1 & -\lambda^2+1 & 1-1 \\ 1+1 & 1+1 & -\lambda^2-1\end{array}\right|=0 \\ & \Rightarrow\left|\begin{array}{ccc}-\lambda^2+1 & 2 & 0 \\ 2 & -\lambda^2+1 & 0 \\ 2 & 2 & -\lambda^2-1\end{array}\right|=0 \\ & \Rightarrow\left(-\lambda^2-1\right)\left\{\left(-\lambda^2+1\right)^2-4\right\}=0 \\ & \Rightarrow-\lambda^2=1,-\lambda^2+1=2 \text { or }-\lambda^2+1=-2 \\ & \Rightarrow \lambda^2=-1 \text { or } \lambda^2=3 \\ & \Rightarrow \lambda= \pm \sqrt{3}\left[\because \lambda^2=-1 \text { not possible }\right] \\ & \end{aligned}$
$\begin{aligned} & \left|\begin{array}{ccc}-\lambda^2+1 & 1+1 & 1-1 \\ 1+1 & -\lambda^2+1 & 1-1 \\ 1+1 & 1+1 & -\lambda^2-1\end{array}\right|=0 \\ & \Rightarrow\left|\begin{array}{ccc}-\lambda^2+1 & 2 & 0 \\ 2 & -\lambda^2+1 & 0 \\ 2 & 2 & -\lambda^2-1\end{array}\right|=0 \\ & \Rightarrow\left(-\lambda^2-1\right)\left\{\left(-\lambda^2+1\right)^2-4\right\}=0 \\ & \Rightarrow-\lambda^2=1,-\lambda^2+1=2 \text { or }-\lambda^2+1=-2 \\ & \Rightarrow \lambda^2=-1 \text { or } \lambda^2=3 \\ & \Rightarrow \lambda= \pm \sqrt{3}\left[\because \lambda^2=-1 \text { not possible }\right] \\ & \end{aligned}$
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