For a, b ∈ Z and | a - b | ≤ 10 , let the angle between the plane...

Given,

For $a, b \in Z$ and $| a - b | \le 10$, let the angle between the plane $P: ax + y - z = b$ and the line $L: x - 1 = a - y = z + 1$ be $\cos^{- 1}\left(\dfrac{1}{3}\right)$

Now rewriting the equation of line we get,

$L: \dfrac{x - 1}{1} = \dfrac{y - a}{- 1} = \dfrac{z + 1}{1}$

Now let, $\theta = \cos^{- 1}\left(\dfrac{1}{3}\right) = \sin^{- 1}\left(\dfrac{2\sqrt{2}}{3}\right)$

Now we know that,

Angle between plane and line is given by,

$\cos \left(90 ° - \theta\right) = \dfrac{a + 1\left(- 1\right) + \left(- 1\right)1}{\sqrt{{a}^{2} + 2} \cdot \sqrt{3}}$

$⇒ \dfrac{a - 2}{\sqrt{{a}^{2} + 2}} = \dfrac{2\sqrt{2}}{\sqrt{3}}$

$⇒ 3{\left(a - 2\right)}^{2} = 8\left({a}^{2} + 2\right)$

$⇒ 3\left({a}^{2} + 4 - 4a\right) = 8\left({a}^{2} + 2\right)$

$⇒ 5{a}^{2} + 12a + 4 = 0$

So, $a = - \dfrac{2}{5}, a = - 2$

Hence, $a = - 2 ( ∵ a \in Z )$

So, the equation of plane will be,

$P: - 2x + y - z - b = 0$

Now given, distance of point $\left(6, - 6, 4\right)$ from plane is$= 3\sqrt{6}$

$⇒ \left|\dfrac{- 12 - 6 - 4 - b}{\sqrt{{2}^{2} + {1}^{2} + {1}^{2}}}\right| = 3\sqrt{6}$

$⇒ | 22 + b | = 18$

$⇒ b = - 4, - 40$

So, $b = - 4 ( ∵ | a - b | \le 10 )$

Hence, ${a}^{4} + {b}^{2} = 16 + 16 = 32$