For a,b$\in$Z and $|\mathrm{a}-\mathrm{b}|\le 10$, let the angle between the plane $P : ax + y - z = b$and the line  $\mathrm{\ell }:\mathrm{x}-1=\mathrm{a}-\mathrm{y}=\mathrm{z}+1 \mathrm{be} {\cos }^{-1}\left(\frac{1}{3}\right)$. If the distance of the point (6, -6, 4) from the plane P is 3$\sqrt{6}$, then a⁴ + b² is equal to

Plane ax + y - z = b
Line $\frac{\mathrm{x}-1}{1}=\frac{\mathrm{y}-a}{-1}=\frac{\mathrm{z}+1}{1}$
$\begin{array}{l}\theta ={\cos }^{-1}\left(\frac{1}{3}\right)={\sin }^{-1}\left(\frac{2\sqrt{2}}{3}\right)\\ \cos \left({90}^{\circ }-\theta \right)=\frac{a+1(-1)+(-1)\left(1\right)}{\sqrt{a^2+2}\cdot \sqrt{3}}=\frac{2\sqrt{2}}{3}\\ \Rightarrow \frac{a-2}{\sqrt{a^2+2}}=\frac{2\sqrt{2}}{\sqrt{3}}\end{array}$
$$\begin{gathered} \sqrt{3}(a-2)=2\sqrt{2}\sqrt{a^2+2} \\ 3(a^2-4a+4)=8(a^2+2) \\ 3a^2-12a+12=8a^2+16 \\ 5a^2+12a+4=0 \\ (5a+2)(a+2)=0 \\ \therefore a=-2, -\frac{2}{5}(\because a\in Z) \end{gathered}$$
$\begin{array}{l}\left|\frac{6a-6-4-b}{\sqrt{a^2+2}}\right|=3\sqrt{6}\\ \Rightarrow |22+b|=18\\ \Rightarrow b=-4,-18-22=-40\\ \therefore b=-4(\because |a-b|\le 10)\\ \therefore a^4+b^2=16+16=32\end{array}$