1. Key Concepts and Formulas

• Family of Planes: The equation of any plane passing through the line of intersection of two planes $P_1: A_1x + B_1y + C_1z + D_1 = 0$ and $P_2: A_2x + B_2y + C_2z + D_2 = 0$ is given by the linear combination: $P_1 + \lambda P_2 = 0$ where $\lambda$ is a scalar constant. This formula is crucial because rotating a plane about its line of intersection with another plane means the new plane still passes through that same line.
• Perpendicularity of Planes: Two planes are perpendicular if and only if their normal vectors are orthogonal. If $\vec{n_1}$ is the normal vector of the first plane and $\vec{n_2}$ is the normal vector of the second plane, then they are perpendicular if: $\vec{n_1} \cdot \vec{n_2} = 0$ A rotation by $\frac{\pi}{2}$ (or $90^\circ$) about a line within a plane ensures that the original plane and its new position are perpendicular.
• Distance from a Point to a Plane: The distance $D$ of a point $(x_0, y_0, z_0)$ from a plane $Ax + By + Cz + D_{plane} = 0$ is given by the formula: $D = \frac{|Ax_0 + By_0 + Cz_0 + D_{plane}|}{\sqrt{A^2 + B^2 + C^2}}$

2. Step-by-Step Solution

Step 1: Formulate the Equation of the New Plane

• What we are doing: We are determining the general algebraic form of the new plane after rotation.
• Why we are doing it: Since the plane P is rotated about its line of intersection with plane Q, the new plane will still pass through this line. We use the family of planes concept to represent all such planes.
• The original plane P is $P_1: 4x - y + z - 10 = 0$. Its normal vector is $\vec{n_1} = \langle 4, -1, 1 \rangle$.
• The plane it intersects with is $P_2: x + y - z - 4 = 0$.
• The equation of the new plane (let's call it $P'$) must be of the form $P_1 + \lambda P_2 = 0$: $(4x - y + z - 10) + \lambda(x + y - z - 4) = 0$
• To easily identify the normal vector of $P'$, we rearrange this equation by collecting coefficients of $x, y, z$: $(4 + \lambda)x + (-1 + \lambda)y + (1 - \lambda)z - (10 + 4\lambda) = 0$
• The normal vector of the new plane $P'$ is $\vec{n'} = \langle 4+\lambda, \lambda-1, 1-\lambda \rangle$.

Step 2: Apply the Perpendicularity Condition

• What we are doing: We are using the given rotation angle to establish a relationship between the normal vectors of the original and new planes.
• Why we are doing it: The problem states that plane P is rotated by an angle of $\frac{\pi}{2}$ (or $90^\circ$) about its line of intersection. This means the original plane P and the new plane P' are perpendicular to each other. For two planes to be perpendicular, their normal vectors must be orthogonal.
• The normal vector of the original plane $P_1$ is $\vec{n_1} = \langle 4, -1, 1 \rangle$.
• The normal vector of the new plane $P'$ is $\vec{n'} = \langle 4+\lambda, \lambda-1, 1-\lambda \rangle$.
• For perpendicularity, their dot product must be zero: $\vec{n_1} \cdot \vec{n'} = 0$

Step 3: Solve for the Parameter $\lambda$

• What we are doing: We are calculating the dot product and solving the resulting equation to find the specific value of $\lambda$.
• Why we are doing it: Finding $\lambda$ will allow us to determine the unique equation of the new plane.
• Substitute the components of the normal vectors into the dot product equation: $(4)(4+\lambda) + (-1)(\lambda-1) + (1)(1-\lambda) = 0$
• Expand and simplify the equation: $(16 + 4\lambda) + (-\lambda + 1) + (1 - \lambda) = 0$ $16 + 4\lambda - \lambda + 1 + 1 - \lambda = 0$
• Combine constant terms and $\lambda$ terms: $(16 + 1 + 1) + (4\lambda - \lambda - \lambda) = 0$ $18 + 2\lambda = 0$
• Solve for $\lambda$: $2\lambda = -18$ $\lambda = -9$

Step 4: Determine the Equation of the New Plane

• What we are doing: We are substituting the value of $\lambda$ back into the general equation of the new plane.
• Why we are doing it: This gives us the specific equation of the plane P' after rotation.
• Substitute $\lambda = -9$ into the general equation of $P'$: $(4 + \lambda)x + (-1 + \lambda)y + (1 - \lambda)z - (10 + 4\lambda) = 0$ $(4 + (-9))x + (-1 + (-9))y + (1 - (-9))z - (10 + 4(-9)) = 0$ $(-5)x + (-10)y + (1 + 9)z - (10 - 36) = 0$ $-5x - 10y + 10z - (-26) = 0$ $-5x - 10y + 10z + 26 = 0$
• For convenience, we can multiply the entire equation by -1: $5x + 10y - 10z - 26 = 0$ This is the equation of the new plane.

Step 5: Calculate the Distance from the Point to the New Plane

• What we are doing: We are applying the distance formula to find $\alpha$.
• Why we are doing it: The problem asks for the distance $\alpha$ of the point $(2,3,-4)$ from the new plane.
• The point is $(x_0, y_0, z_0) = (2,3,-4)$.
• The new plane is $5x + 10y - 10z - 26 = 0$, so $A=5, B=10, C=-10, D_{plane}=-26$.
• Using the distance formula: $\alpha = \frac{|(5)(2) + (10)(3) + (-10)(-4) - 26|}{\sqrt{(5)^2 + (10)^2 + (-10)^2}}$
• Calculate the numerator: $|10 + 30 + 40 - 26| = |80 - 26| = |54| = 54$
• Calculate the denominator: $\sqrt{25 + 100 + 100} = \sqrt{225} = 15$
• So, the distance $\alpha$ is: $\alpha = \frac{54}{15}$
• Simplify the fraction by dividing both numerator and denominator by 3: $\alpha = \frac{18}{5}$

Step 6: Compute the Final Value

• What we are doing: We are performing the final calculation requested by the problem statement.
• Why we are doing it: The question asks for the value of $35\alpha$.
• $35\alpha = 35 \times \frac{18}{5}$
• Simplify the expression: $35\alpha = (7 \times 5) \times \frac{18}{5}$ $35\alpha = 7 \times 18$ $35\alpha = 126$

3. Common Mistakes & Tips

• Sign Errors: Be meticulous with signs, especially when substituting coordinates into the distance formula and when calculating the dot product. A single sign error can lead to an incorrect $\lambda$ or an incorrect distance.
• Absolute Value: Remember to take the absolute value of the numerator in the distance formula, as distance must always be non-negative.
• Algebraic Simplification: Double-check calculations when solving for $\lambda$ and simplifying fractions. Careless arithmetic is a common pitfall.

4. Summary

This problem involves applying fundamental concepts of 3D geometry. We first used the family of planes concept to represent the new plane, incorporating an unknown parameter $\lambda$. The crucial information about the $\frac{\pi}{2}$ rotation allowed us to establish that the original and new planes are perpendicular, leading to a condition on their normal vectors (dot product is zero). Solving this condition yielded the value of $\lambda$, which in turn defined the specific equation of the new plane. Finally, we calculated the distance from the given point to this new plane using the standard distance formula and then computed the required value of $35\alpha$.

The final answer is $\boxed{126}$, which corresponds to option (A).