JEE Main 2022 — Three Dimensional Geometry Question with Solution

Plotting the rough graph of the given data we have,

Let $M$ be the mid-point of $PQ$

$\therefore M=\left(3\lambda +6,2\lambda +1,3\lambda +2\right)$

Now, $\overrightarrow{PM}=\left(3\lambda +5\right)\hat{i}+\left(2\lambda -1\right)\hat{j}+\left(3\lambda -1\right)\hat{k}$

$\because \overrightarrow{PM}\perp \left(3\hat{i}+2\hat{j}+3\hat{k}\right)$

$\therefore 3\left(3\lambda +5\right)+2\left(2\lambda -1\right)+3\left(3\lambda -1\right)=0$

$\lambda =\frac{-5}{11}$

$\therefore M\left(\frac{51}{11},\frac{1}{11},\frac{7}{11}\right)$

Since $R$ is mid-point of $PM$

So by using midpoint formula in $P \& M$ we get,

$R=\left(\frac{62}{22},\frac{23}{22},\frac{40}{22}\right)$, now comparing with $R\left(\alpha ,\beta ,\gamma \right)$

We get, $\alpha =\frac{62}{22}, \beta =\frac{23}{22} \& \gamma =\frac{40}{22}$

So, $22\left(\alpha +\beta +\gamma \right)=125$