1. Key Concepts and Formulas

This problem involves concepts from 3D Geometry:

• Angle between a Plane and a Line: If a plane has a normal vector $\vec{n} = (A,B,C)$ and a line has a direction vector $\vec{d} = (l,m,n)$, the angle $\theta$ between the plane and the line is given by: $\sin \theta = \frac{|\vec{n} \cdot \vec{d}|}{||\vec{n}|| \cdot ||\vec{d}||}$ where $\vec{n} \cdot \vec{d}$ is the dot product, $||\vec{n}|| = \sqrt{A^2+B^2+C^2}$ is the magnitude of the normal vector, and $||\vec{d}|| = \sqrt{l^2+m^2+n^2}$ is the magnitude of the direction vector.
• Distance of a Point from a Plane: The distance $D$ of a point $(x_1, y_1, z_1)$ from a plane $Ax+By+Cz+D'=0$ is given by: $D = \frac{|Ax_1+By_1+Cz_1+D'|}{\sqrt{A^2+B^2+C^2}}$

2. Step-by-Step Solution

Step 1: Determine $\sin \theta$ from the given angle. The angle between the plane P and the line $l$ is given as $\theta = \cos^{-1}\left(\frac{1}{3}\right)$. This means $\cos \theta = \frac{1}{3}$. To use the angle formula for a plane and a line, we need $\sin \theta$. We use the identity $\sin^2 \theta + \cos^2 \theta = 1$. Since $\theta$ is an angle between a line and a plane, $0 \le \theta \le \frac{\pi}{2}$, so $\sin \theta$ must be positive. $\sin \theta = \sqrt{1 - \cos^2 \theta} = \sqrt{1 - \left(\frac{1}{3}\right)^2} = \sqrt{1 - \frac{1}{9}} = \sqrt{\frac{8}{9}} = \frac{2\sqrt{2}}{3}$

Step 2: Extract the normal vector of the plane and the direction vector of the line. The equation of the plane P is $ax+y-z=b$. The normal vector to the plane P, $\vec{n}$, is obtained from the coefficients of $x, y, z$: $\vec{n} = (a, 1, -1)$ The equation of the line $l$ is $x-1 = a-y = z+1$. To find its direction vector, we must write the line in its standard symmetric form, $\frac{x-x_0}{l} = \frac{y-y_0}{m} = \frac{z-z_0}{n}$: $x-1 = \frac{y-a}{-1} = \frac{z-(-1)}{1}$ From this form, the direction vector of the line $l$, $\vec{d}$, is: $\vec{d} = (1, -1, 1)$

Step 3: Apply the angle formula and solve for 'a'. Using the formula for the angle between a plane and a line: $\sin \theta = \frac{|\vec{n} \cdot \vec{d}|}{||\vec{n}|| \cdot ||\vec{d}||}$ First, calculate the dot product $\vec{n} \cdot \vec{d}$: $\vec{n} \cdot \vec{d} = (a)(1) + (1)(-1) + (-1)(1) = a - 1 - 1 = a - 2$ Next, calculate the magnitudes of the vectors: $||\vec{n}|| = \sqrt{a^2 + 1^2 + (-1)^2} = \sqrt{a^2 + 1 + 1} = \sqrt{a^2 + 2}$ $||\vec{d}|| = \sqrt{1^2 + (-1)^2 + 1^2} = \sqrt{1 + 1 + 1} = \sqrt{3}$ Substitute these values into the $\sin \theta$ formula from Step 1: $\frac{2\sqrt{2}}{3} = \frac{|a-2|}{\sqrt{a^2+2} \cdot \sqrt{3}}$ To solve for $a$, we rearrange and square both sides: $2\sqrt{2}\sqrt{3} \sqrt{a^2+2} = 3|a-2|$ $2\sqrt{6} \sqrt{a^2+2} = 3|a-2|$ Squaring both sides (recall $|X|^2 = X^2$): $(2\sqrt{6})^2 (\sqrt{a^2+2})^2 = (3(a-2))^2$ $(4 \cdot 6)(a^2+2) = 9(a-2)^2$ $24(a^2+2) = 9(a^2 - 4a + 4)$ Divide by 3 to simplify: $8(a^2+2) = 3(a^2 - 4a + 4)$ $8a^2 + 16 = 3a^2 - 12a + 12$ Rearrange into a quadratic equation: $5a^2 + 12a + 4 = 0$ Factor the quadratic equation: $5a^2 + 10a + 2a + 4 = 0$ $5a(a+2) + 2(a+2) = 0$ $(5a+2)(a+2) = 0$ This yields two possible values for $a$: $a = -\frac{2}{5} \quad \text{or} \quad a = -2$ The problem states that $a \in \mathbb{Z}$ (a is an integer). Therefore, we must choose $a = -2$.

Step 4: Use the distance formula to find 'b'. With $a=-2$, the equation of the plane P becomes: $-2x + y - z = b \quad \text{or} \quad -2x + y - z - b = 0$ The distance of the point $(6, -6, 4)$ from this plane is given as $3\sqrt{6}$. Using the distance formula $D = \frac{|Ax_1+By_1+Cz_1+D'|}{\sqrt{A^2+B^2+C^2}}$: Here, $(x_1, y_1, z_1) = (6, -6, 4)$, and from the plane equation, $A=-2, B=1, C=-1, D'=-b$. $3\sqrt{6} = \frac{|(-2)(6) + (1)(-6) + (-1)(4) - b|}{\sqrt{(-2)^2 + 1^2 + (-1)^2}}$ $3\sqrt{6} = \frac{|-12 - 6 - 4 - b|}{\sqrt{4 + 1 + 1}}$ $3\sqrt{6} = \frac{|-22 - b|}{\sqrt{6}}$ Multiply both sides by $\sqrt{6}$: $3\sqrt{6} \cdot \sqrt{6} = |-22 - b|$ $3 \cdot 6 = |-(22 + b)|$ $18 = |22 + b|$ This absolute value equation gives two possibilities for $22+b$: $22+b = 18 \quad \text{or} \quad 22+b = -18$ Solving for $b$: $b = 18 - 22 \implies b = -4$ $b = -18 - 22 \implies b = -40$

Step 5: Apply the condition $|a-b| \leq 10$ to choose the correct 'b'. We have $a = -2$. We check which value of $b$ satisfies the condition $|a-b| \leq 10$.

• Case 1: $b = -4$ $|a-b| = |-2 - (-4)| = |-2 + 4| = |2| = 2$ Since $2 \leq 10$, this value of $b=-4$ is valid.

• Case 2: $b = -40$ $|a-b| = |-2 - (-40)| = |-2 + 40| = |38| = 38$ Since $38 \not\leq 10$, this value of $b=-40$ is not valid.

Therefore, the unique correct values are $a=-2$ and $b=-4$.

Step 6: Calculate $a^4+b^2$. Substitute the determined values of $a$ and $b$ into the expression $a^4+b^2$: $a^4+b^2 = (-2)^4 + (-4)^2$ $= 16 + 16$ $= 32$

3. Common Mistakes & Tips

• Angle Formula: Remember to use $\sin \theta$ for the angle between a plane and a line, not $\cos \theta$. The angle between the normal vector and the line's direction vector is complementary to the angle between the plane and the line.
• Line Equation Form: Always convert the line equation into the standard symmetric form to correctly identify the direction vector. Pay close attention to signs, especially for terms like $a-y$.
• Absolute Values: Absolute value equations always lead to two cases. Make sure to consider both and use all given conditions (like $|a-b| \leq 10$ and $a,b \in \mathbb{Z}$) to filter for the correct solution.
• Integer Constraints: Do not forget to apply integer constraints ($a,b \in \mathbb{Z}$) to filter out non-integer solutions at each relevant step.

4. Summary

This problem required a systematic application of 3D geometry formulas. We first converted the given angle $\cos^{-1}(1/3)$ to $\sin \theta = \frac{2\sqrt{2}}{3}$, which is needed for the angle between a plane and a line. We then extracted the normal vector of the plane and the direction vector of the line, being careful with the line's symmetric form. Applying the angle formula led to a quadratic equation for $a$, from which we found $a=-2$ (since $a$ must be an integer). Next, using the distance formula from a point to the plane, we found two possible integer values for $b$. Finally, the condition $|a-b| \leq 10$ uniquely determined $b=-4$. Substituting these values into $a^4+b^2$ gave the final result.

The final answer is $\boxed{48}$, which corresponds to option (A).