3D Geometry 3 Question 16
####16. 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
(2019 Main, 12 Jan II)
(a) ${\sqrt{3},-\sqrt{3}}$
(b) ${3,-3}$
(c) ${1,-1}$
(d) ${\sqrt{3}}$
Show Answer
Answer:
(a)
Solution:
- According to the question points $\left(-\lambda^{2}, 1,1\right),\left(1,-\lambda^{2}, 1\right)$ and $\left(1,1,-\lambda^{2}\right)$ are coplanar with the point $(-1,-1,1)$, so
$ \begin{aligned} & \left|\begin{array}{ccc} 1-\lambda^{2} & 2 & 0 \\ 2 & 1-\lambda^{2} & 0 \\ 2 & 2 & -\lambda^{2}-1 \end{array}\right|=0 \\ & \left|\begin{array}{ccc} x_{2}-x_{1} & y_{2}-y_{1} & z_{2}-z_{1} \\ x_{3}-x_{1} & y_{3}-y_{1} & z_{3}-z_{1} \\ x_{4}-x_{1} & y_{4}-y_{1} & z_{4}-z_{1} \end{array}\right|=0 \\ & \Rightarrow \quad\left(-1-\lambda^{2}\right)\left[\left(1-\lambda^{2}\right)^{2}-4\right]=0 \\ & \Rightarrow\left(1+\lambda^{2}\right)\left[\left(1-\lambda^{2}-2\right)\left(1-\lambda^{2}+2\right)\right]=0 \\ & \Rightarrow \quad\left(1+\lambda^{2}\right)^{2}\left(3-\lambda^{2}\right)=0 \\ & \Rightarrow \quad \lambda^{2}=3 \quad\left[\because 1+\lambda^{2} \neq 0 \forall \lambda \in R\right] \\ & \Rightarrow \quad \lambda= \pm \sqrt{3} \end{aligned} $
