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

We have, $\theta=\cos ^{-1} \frac{1}{3}$ <br/><br/>$$ \begin{aligned} & \Rightarrow \cos \theta=\frac{1}{3} \\\\ & \therefore \sin \theta=\sqrt{1-\left(\frac{1}{3}\right)^2}=\sqrt{\frac{8}{9}}=\frac{2 \sqrt{2}}{3} \end{aligned} $$ <br/><br/>The given plane line and are <br/><br/>$a x+y-z=b$ <br/><br/>$x-1=a-y=z+1$ <br/><br/>$$ \begin{aligned} & \therefore \sin \theta=\frac{a \cdot 1+(1)(-1)+(-1)(1)}{\sqrt{a^2+1^2+1^2} \sqrt{1^2+1^2+1^2}} \\\\ & \Rightarrow \frac{a-1-1}{\sqrt{a^2+2} \sqrt{3}}=\frac{2 \sqrt{2}}{3} \\\\ & \Rightarrow 3(a-2)=2 \sqrt{6} \sqrt{a^2+2} \\\\ & \Rightarrow 9\left(a^2+4-4 a\right)=24\left(a^2+2\right) \\\\ & \Rightarrow 9 a^2+36-36 a=24 a^2+48 \\\\ & \Rightarrow 15 a^2+36 a+12=0 \\\\ & \Rightarrow 5 a^2+12 a+4=0 \\\\ & \Rightarrow 5 a^2+10 a+2 a+4=0 \\\\ & \Rightarrow 5 a(a+2)+2(a+2)=0 \\\\ & \Rightarrow a=\frac{-2}{5},-2 \end{aligned} $$ <br/><br/>So, $a=-2$ [$\because a \in Z$] <br/><br/>Hence, the eqn. of plane is $-2 x+y-z-b=0$ <br/><br/>$$ \begin{aligned} & \text { Now, } d=\left|\frac{-12-6-4-b}{\sqrt{4+1+1}}\right|=3 \sqrt{6} \\\\ & \Rightarrow|-(b+22)|=18 \\\\ & \Rightarrow b=18-22=-4 \\\\ & \therefore a^4+b^2=(-2)^4+(-4)^2 \\\\ & =16+16=32 \end{aligned} $$