Question

Let $S$ be the set of all real values of $\lambda$ such that plane passing through the points $(-{\lambda }^2,1,1),(1,-{\lambda }^2,1)$ and $(1,1,-{\lambda }^2)$ also passes through the point $(-1,-1,1)$. Then $S$ is equal to :

• A. $\left\{\sqrt{3}\right\}$
• B. $\{1,-1\}$
• C. $\{3,-3\}$
• D. $\{\sqrt{3},-\sqrt{3}\}$

Answer 1

The correct option is D $\{\sqrt{3},-\sqrt{3}\}$

The given points $(-1,-1,1)$, $(-{\lambda }^2,1,1),(1,-{\lambda }^2,1)$ and $(1,1,-{\lambda }^2)$ are coplanar.

$\therefore \left|\begin{array}{ccc}-1+{\lambda }^2& -2& 0\\ -2& -1+{\lambda }^2& 0\\ -2& -2& 1+{\lambda }^2\end{array}\right|=0$

$\Rightarrow (1+{\lambda }^2)\left[\right({\lambda }^2-1{)}^2-4]=0$
$\Rightarrow ({\lambda }^2+1{)}^2({\lambda }^2-3)=0$
​​​​​​​$\Rightarrow {\lambda }^2=3$ or $\lambda =\pm \sqrt{3}$