For a, b in mathbb{Z} and |a - b| leq 10,let | vedclass

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(D) The plane is $P: ax + y - z - b = 0$. The normal vector is $\vec{n} = a\hat{i} + \hat{j} - \hat{k}$.The line $l$ is $\frac{x-1}{1} = \frac{y}{-1}

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$32$

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(D) The plane is $P: ax + y - z - b = 0$. The normal vector is $\vec{n} = a\hat{i} + \hat{j} - \hat{k}$.The line $l$ is $\frac{x-1}{1} = \frac{y}{-1} = \frac{z+1}{1}$. The direction vector is $\vec{v} = \hat{i} - \hat{j} + \hat{k}$.The angle $	heta$ between the line and the plane satisfies $\sin 	heta = \frac{|\vec{v} \cdot \vec{n}|}{|\vec{v}| |\vec{n}|}$.Given $\cos 	heta = \frac{1}{3}$,so $\sin 	heta = \sqrt{1 - (1/3)^2} = \frac{\sqrt{8}}{3} = \frac{2\sqrt{2}}{3}$.$\frac{|a(1) + 1(-1) + (-1)(1)|}{\sqrt{1^2 + (-1)^2 + 1^2} \sqrt{a^2 + 1^2 + (-1)^2}} = \frac{2\sqrt{2}}{3} \implies \frac{|a - 2|}{\sqrt{3} \sqrt{a^2 + 2}} = \frac{2\sqrt{2}}{3}$.Squaring both sides: $\frac{(a-2)^2}{3(a^2+2)} = \frac{8}{9} \implies 3(a^2 - 4a + 4) = 8(a^2 + 2) \implies 3a^2 - 12a + 12 = 8a^2 + 16$.$5a^2 + 12a + 4 = 0 \implies (5a + 2)(a + 2) = 0$. Since $a \in \mathbb{Z}$,$a = -2$.Distance from $(6, -6, 4)$ to $-2x + y - z - b = 0$ is $\frac{|-2(6) + (-6) - 4 - b|}{\sqrt{(-2)^2 + 1^2 + (-1)^2}} = 3\sqrt{6}$.$\frac{|-12 - 6 - 4 - b|}{\sqrt{6}} = 3\sqrt{6} \implies |-22 - b| = 18$.$b + 22 = 18 \implies b = -4$ or $b + 22 = -18 \implies b = -40$.Given $|a - b| \leq 10$,for $a = -2$: $|-2 - (-4)| = 2 \leq 10$ (Valid),$|-2 - (-40)| = 38 
ot\leq 10$.Thus $a = -2, b = -4$. $a^4 + b^2 = (-2)^4 + (-4)^2 = 16 + 16 = 32$.

For $a, b \in \mathbb{Z}$ and $|a - b| \leq 10$,let the angle between the plane $P: ax + y - z = b$ and the line $l: x - 1 = \frac{-y}{1} = z + 1$ 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

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