For and , let the angle between the plane and the line be . If the distance of the point from the plane is , then is equal to

32 Explanation: We have, The given plane line and are \begin{aligned} & \Rightarrow3(a-2)=2\sqrt{6}\sqrt{a^{2}+2} \\ & \Rightarrow9(a^{2}+4-4a)=24(a^{2}+2) \\ & \Rightarrow9a^{2}+36-36a=24a^{2}+48 \\ & \Rightarrow15a^{2}+36a+12=0 \\ & \Rightarrow5a^{2}+12a+4=0 \\ & \Rightarrow5a^{2}+10a+2a+4=0 \\ & \Rightarrow5a\left(a+2\right)+2\left(a+2\right)=0 \\ & \Rightarrow a=\frac{-2}{5},-2 \\ & \mathrm{So},a=-2 & \because a\in\mathbb{Z} \\ & \mathrm{Hence,~the~eqn.~of~plane~is~-2x~+~y-z-b=0} \\ & \mathrm{Now,}d=\left|\frac{-12-6-4-b}{\sqrt{4+1+1}}\right|=3\sqrt{6} \\ & \Rightarrow|-(b+22)|=18 \\ & \Rightarrow b=18-22=-4 \\ & \therefore a^{4}+b^{2}=\left(-2\right)^{4}+\left(-4\right)^{2} \\ & =16+16=32 \end{aligned}

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JEE 2023 — Mathematics PYQ

JEE | Mathematics | 2023

For $a, b \in Z$ and $∣ a - b ∣ \leq 10$ , let the angle between the plane $P : a x + y - z = b$ and the line $l : x - 1 = a - y = z + 1$ be $cos^{- 1} (\frac{1}{3})$ . If the distance of the point $(6, - 6, 4)$ from the plane $P$ is $36$ , then $a^{4} + b^{2}$ is equal to

JEE 2023 — Mathematics PYQ

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