Question:
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:
Correct Option: , 2
Solution:
All four points are coplaner so
$\left|\begin{array}{ccc}1-\lambda^{2} & 2 & 0 \\ 2 & -\lambda^{2}+1 & 0 \\ 2 & 2 & -\lambda^{2}-1\end{array}\right|=0$
$\left(\lambda^{2}+1\right)^{2}\left(3-\lambda^{2}\right)=0$
$\lambda=\pm \sqrt{3}$
Click here to get exam-ready with eSaral
For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.