uj.s3.ap-south-1.amazonaws.com/pyq-question-images/batch-001/7059dadb9dd0.webp" loading="lazy" style="max-width:100%;height:auto;display:block;margin:0px auto;max-height:40vh;vertical-align:baseline" alt="JEE Main 2023 (Online) 10th April Evening Shift Mathematics - 3D Geometry Question 123 English Explanation">

We have, equation of plane $P$ is $x-y+2 z+3=0$

Perpendicular distance from $A$ to a plane $P$ i.e. N

$=\frac{|4-3+2(1)+3|}{\sqrt{1^2+(-1)^2+2^2}}$

$B(5, \alpha, \beta)$ lies on the plane $P$, so

$5-\alpha+2 \beta+3=0 \Rightarrow \alpha=2 \beta+8$ .........(i)

Direction ratio of $A N$ is $<1,-1,2>$

Equation of $A N$ is $\frac{x-4}{1}=\frac{y-3}{-1}=\frac{z-1}{2}$

Co-ordinate of $N: \frac{x-4}{1}=\frac{y-3}{-1}=\frac{z-1}{2}=\frac{-(4-3+2+3)}{1+1+4}$ $=-1$

$\Rightarrow x=3, y=4, z=-1$

$\therefore$ Co-ordinate of $N$ is $(3,4,-1)$

$$ \begin{aligned} & \text { Also, } \operatorname{Area of}(\triangle \mathrm{ABN})=3 \sqrt{2} \\\\ & \Rightarrow \frac{1}{2} \times \mathrm{AN} \times \mathrm{BN}=3 \sqrt{2} \\\\ & \Rightarrow \frac{1}{2} \times \sqrt{6} \times \mathrm{BN}=3 \sqrt{2} \Rightarrow \mathrm{BN}=2 \sqrt{3} \\\\ & \text { or } \mathrm{BN}^2=12 \end{aligned} $$

$$ \begin{aligned} & \Rightarrow(5-3)^2+(\alpha-4)^2+(\beta+1)^2=12 \\\\ & 4+(2 \beta+4)^2+(\beta+1)^2=12(\text { using eqn (i)) } \\\\ & \Rightarrow 4 \beta^2+16+16 \beta+\beta^2+1+2 \beta=8 \\\\ & \Rightarrow 5 \beta^2+18 \beta+9=0 \end{aligned} $$

$$ \begin{aligned} & \Rightarrow(5 \beta+3)(\beta+3)=0 \\\\ & \Rightarrow \beta=-3 \left[\because \beta \in z \text {, so rejecting } \beta=\frac{-3}{5}\right]\\\\ &\text { and } \alpha=2 \beta+8=2(-3)+8=2\\\\ & \text { also } \alpha^2+\beta^2+\alpha \beta=9+4-6=7 \end{aligned} $$

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