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 $$L: x - 1 = a - y = z + 1$$ 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 are given: $$a, b \in \mathbb{Z}$$, $$|a - b| \leq 10$$, the angle between plane $$P: ax + y - z = b$$ and line $$L: x - 1 = a - y = z + 1$$ is $$\cos^{-1}\left(\frac{1}{3}\right)$$, and the distance of $$(6, -6, 4)$$ from plane $$P$$ is $$3\sqrt{6}$$. Find $$a^4 + b^2$$.

First, we find the direction vector of line $$L$$.

The line is $$x - 1 = a - y = z + 1$$, which can be written as:

$$\frac{x - 1}{1} = \frac{y - a}{-1} = \frac{z + 1}{1}$$

So the direction vector of $$L$$ is $$\vec{d} = (1, -1, 1)$$.

Next, we find the normal vector of plane $$P$$.

The plane $$ax + y - z = b$$ has normal vector $$\vec{n} = (a, 1, -1)$$.

From this, we use the angle between line and plane.

If $$\alpha$$ is the angle between a line and a plane, then:

$$\sin \alpha = \frac{|\vec{n} \cdot \vec{d}|}{|\vec{n}||\vec{d}|}$$

Given $$\alpha = \cos^{-1}\left(\frac{1}{3}\right)$$, we have:

$$\cos \alpha = \frac{1}{3}$$, so $$\sin \alpha = \sqrt{1 - \frac{1}{9}} = \frac{2\sqrt{2}}{3}$$

Now compute:

$$\vec{n} \cdot \vec{d} = a(1) + (1)(-1) + (-1)(1) = a - 2$$

$$|\vec{n}| = \sqrt{a^2 + 1 + 1} = \sqrt{a^2 + 2}$$

$$|\vec{d}| = \sqrt{1 + 1 + 1} = \sqrt{3}$$

So:

$$\frac{|a - 2|}{\sqrt{3(a^2 + 2)}} = \frac{2\sqrt{2}}{3}$$

Now, we solve for $$a$$.

Squaring both sides:

$$\frac{(a - 2)^2}{3(a^2 + 2)} = \frac{8}{9}$$

Cross-multiplying:

$$9(a - 2)^2 = 24(a^2 + 2)$$

$$9a^2 - 36a + 36 = 24a^2 + 48$$

$$15a^2 + 36a + 12 = 0$$

$$5a^2 + 12a + 4 = 0$$

Using the quadratic formula:

$$a = \frac{-12 \pm \sqrt{144 - 80}}{10} = \frac{-12 \pm 8}{10}$$

$$a = -\frac{2}{5}$$ or $$a = -2$$

Since $$a \in \mathbb{Z}$$, we get $$a = -2$$.

Then, we find $$b$$ using the distance condition.

Distance of $$(6, -6, 4)$$ from plane $$-2x + y - z = b$$:

$$\frac{|-2(6) + (-6) - 4 - b|}{\sqrt{4 + 1 + 1}} = 3\sqrt{6}$$

$$\frac{|-22 - b|}{\sqrt{6}} = 3\sqrt{6}$$

$$|22 + b| = 18$$

So $$b = -4$$ or $$b = -40$$.

Continuing, we apply the constraint $$|a - b| \leq 10$$.

For $$b = -4$$: $$|-2 - (-4)| = 2 \leq 10$$ ✓

For $$b = -40$$: $$|-2 - (-40)| = 38 > 10$$ ✗

So $$b = -4$$.

It follows that we compute the answer.

$$a^4 + b^2 = (-2)^4 + (-4)^2 = 16 + 16 = 32$$

The correct answer is Option A: 32.