# Ley O be the origin.

Question:

Let $\mathrm{O}$ be the origin. Let $\overrightarrow{\mathrm{OP}}=\mathrm{xi}+\hat{\mathrm{i}}+\hat{\mathrm{j}}-\hat{\mathrm{k}}$ and $\overrightarrow{\mathrm{OQ}}=-\hat{\mathrm{i}}+2 \hat{\mathrm{j}}+3 x \hat{\mathrm{k}}, x, y \in R, x>0$, be such that $|\overrightarrow{\mathrm{PQ}}|=\sqrt{20}$ and the vector $\overrightarrow{\mathrm{OP}}$ is perpendicular

to $\overrightarrow{\mathrm{OQ}}$. If $\overrightarrow{\mathrm{OR}}=3 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}-7 \hat{\mathrm{k}}, \mathrm{z} \in \mathrm{R}$, is coplanar with $\overrightarrow{\mathrm{OP}}$ and $\overrightarrow{\mathrm{OQ}}$, then the value of $x^{2}+y^{2}+z^{2}$ is equal to

1. 7

2. 9

3. 2

4. 1

Correct Option: , 2

Solution:

$\overrightarrow{\mathrm{OP}} \perp \overrightarrow{\mathrm{OQ}}$

$\Rightarrow-\mathrm{x}+2 \mathrm{y}-3 \mathrm{x}=0$

$\Rightarrow \mathrm{y}=2 \mathrm{x}$..(1)

$|\overrightarrow{\mathrm{PQ}}|^{2}=20$

$\Rightarrow(\mathrm{x}+1)^{2}+(\mathrm{y}-2)^{2}+(1+3 \mathrm{x})^{2}=20$

$\Rightarrow \mathrm{x}=1$

$\overrightarrow{\mathrm{OP}}, \overrightarrow{\mathrm{OO}}, \overrightarrow{\mathrm{OR}}$ are coplanar.

$\Rightarrow\left|\begin{array}{ccc}x & y & -1 \\ -1 & 2 & 3 x \\ 3 & z & -7\end{array}\right|=0$

$\Rightarrow\left|\begin{array}{ccc}1 & 2 & -1 \\ -1 & 2 & 3 \\ 3 & \mathrm{z} & -7\end{array}\right|=0$

$\Rightarrow 1(-14-3 z)-2(7-9)-1(-z-6)=0$

$\Rightarrow z=-2$

$\therefore x^{2}+y^{2}+z^{2}=1+4+4=9$ Option $(2)$