1. Key Concepts and Formulas

• Equation of a Plane: A plane passing through a point $(x_0, y_0, z_0)$ with a normal vector $\vec{n} = \langle A, B, C \rangle$ has the equation $A(x-x_0) + B(y-y_0) + C(z-z_0) = 0$.
• Normal Vector to a Plane:
  • If a plane contains two non-parallel lines with direction vectors $\vec{v}_1$ and $\vec{v}_2$, its normal vector is parallel to their cross product: $\vec{n} \propto \vec{v}_1 \times \vec{v}_2$.
  • If a line with direction vector $\vec{v}$ lies in a plane, the normal vector of the plane, $\vec{n}$, is perpendicular to $\vec{v}$ (i.e., $\vec{n} \cdot \vec{v} = 0$).
  • If two planes are perpendicular, their normal vectors are perpendicular (i.e., $\vec{n}_1 \cdot \vec{n}_2 = 0$).
• 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

Our goal is to find the equation of plane P, and then calculate the distance from the given point to this plane. We will achieve this by first finding the normal vector to plane P and a point on it.

Step 1: Determine the Normal Vector of the Second Plane ($P_2$)

The problem states that plane $P_2$ contains two straight lines: $L_2: \frac{x}{2}=\frac{y}{3}=\frac{z}{5}$ and $L_3: \frac{x}{3}=\frac{y}{7}=\frac{z}{8}$.

• What and Why: Since these two lines lie in plane $P_2$, their direction vectors must also lie in $P_2$. The normal vector to $P_2$ must be perpendicular to every vector lying in $P_2$, including the direction vectors of $L_2$ and $L_3$. Therefore, we can find the direction ratios of the normal vector to $P_2$ by taking the cross product of the direction vectors of $L_2$ and $L_3$.

• Math:

  • Direction vector of $L_2$, $\vec{v}_2 = \langle 2, 3, 5 \rangle$.
  • Direction vector of $L_3$, $\vec{v}_3 = \langle 3, 7, 8 \rangle$.
  • Let $\vec{n}_2 = \langle A_2, B_2, C_2 \rangle$ be the normal vector to plane $P_2$. $\vec{n}_2 = \vec{v}_2 \times \vec{v}_3 = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 2 & 3 & 5 \\ 3 & 7 & 8 \end{vmatrix}$ $= \mathbf{i}(3 \times 8 - 5 \times 7) - \mathbf{j}(2 \times 8 - 5 \times 3) + \mathbf{k}(2 \times 7 - 3 \times 3)$ $= \mathbf{i}(24 - 35) - \mathbf{j}(16 - 15) + \mathbf{k}(14 - 9)$ $= -11\mathbf{i} - 1\mathbf{j} + 5\mathbf{k}$
  • Reasoning: The normal vector to plane $P_2$ is $\vec{n}_2 = \langle -11, -1, 5 \rangle$. (We can also use $\langle 11, 1, -5 \rangle$ as it represents the same direction).

Step 2: Determine the Normal Vector of Plane P ($\vec{n}_P$)

Plane P has two defining properties:

1. It contains the straight line $L_1: \frac{x-3}{9}=\frac{y+4}{-1}=\frac{z-7}{-5}$.
2. It is perpendicular to plane $P_2$.

• What and Why:

  • From property 1, the direction vector of $L_1$ must lie in plane P. This means the normal vector of plane P, $\vec{n}_P$, must be perpendicular to the direction vector of $L_1$.
  • From property 2, since plane P is perpendicular to plane $P_2$, their normal vectors must be perpendicular to each other.
  • Therefore, $\vec{n}_P$ must be perpendicular to both $\vec{v}_1$ (direction vector of $L_1$) and $\vec{n}_2$ (normal vector of $P_2$). This implies $\vec{n}_P$ is parallel to their cross product.

• Math:

  • Direction vector of $L_1$, $\vec{v}_1 = \langle 9, -1, -5 \rangle$.
  • Normal vector of $P_2$, $\vec{n}_2 = \langle -11, -1, 5 \rangle$.
  • Let $\vec{n}_P = \langle A, B, C \rangle$ be the normal vector to plane P. $\vec{n}_P = \vec{v}_1 \times \vec{n}_2 = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 9 & -1 & -5 \\ -11 & -1 & 5 \end{vmatrix}$ $= \mathbf{i}((-1)(5) - (-5)(-1)) - \mathbf{j}((9)(5) - (-5)(-11)) + \mathbf{k}((9)(-1) - (-1)(-11))$ $= \mathbf{i}(-5 - 5) - \mathbf{j}(45 - 55) + \mathbf{k}(-9 - 11)$ $= -10\mathbf{i} + 10\mathbf{j} - 20\mathbf{k}$
  • Reasoning: The normal vector to plane P is $\vec{n}_P = \langle -10, 10, -20 \rangle$. We can simplify this by dividing by $-10$ to get $\vec{n}_P = \langle 1, -1, 2 \rangle$. This simpler vector represents the same normal direction.

Step 3: Determine the Equation of Plane P

• What and Why: We have the normal vector to plane P ($\vec{n}_P = \langle 1, -1, 2 \rangle$) and a point lying on the plane. The line $L_1$ lies in plane P, and the equation of $L_1$ is given in symmetric form $\frac{x-3}{9}=\frac{y+4}{-1}=\frac{z-7}{-5}$, which indicates that the point $(3, -4, 7)$ lies on $L_1$, and therefore on plane P. We can now use the point-normal form of the plane equation.

• Math:

  • Normal vector $\vec{n}_P = \langle 1, -1, 2 \rangle$.
  • Point on plane P, $(x_0, y_0, z_0) = (3, -4, 7)$.
  • Equation of plane P: $A(x-x_0) + B(y-y_0) + C(z-z_0) = 0$ $1(x-3) - 1(y-(-4)) + 2(z-7) = 0$ $x-3 - (y+4) + 2z-14 = 0$ $x-3-y-4+2z-14 = 0$ $x-y+2z-21 = 0$
  • Reasoning: The equation of plane P is $x-y+2z-21=0$.

Step 4: Calculate the Distance $d$ of Plane P from the Given Point

• What and Why: We need to find the distance of plane P (equation $x-y+2z-21=0$) from the point $(2, -5, 11)$. We will use the formula for the distance of a point from a plane.

• Math:

  • Plane P: $x-y+2z-21=0$ (so $A=1, B=-1, C=2, D=-21$).
  • Point $(x_1, y_1, z_1) = (2, -5, 11)$.
  • Distance $d = \frac{|Ax_1+By_1+Cz_1+D|}{\sqrt{A^2+B^2+C^2}}$ $d = \frac{|1(2) - 1(-5) + 2(11) - 21|}{\sqrt{1^2 + (-1)^2 + 2^2}}$ $d = \frac{|2 + 5 + 22 - 21|}{\sqrt{1 + 1 + 4}}$ $d = \frac{|7 + 22 - 21|}{\sqrt{6}}$ $d = \frac{|29 - 21|}{\sqrt{6}}$ $d = \frac{|8|}{\sqrt{6}} = \frac{8}{\sqrt{6}}$
  • Reasoning: The distance $d = \frac{8}{\sqrt{6}}$.

Step 5: Calculate $d^2$

• What and Why: The question asks for the value of $d^2$.

• Math: $d^2 = \left(\frac{8}{\sqrt{6}}\right)^2 = \frac{8^2}{(\sqrt{6})^2} = \frac{64}{6}$ $d^2 = \frac{32}{3}$

  • Reasoning: The value of $d^2$ is $\frac{32}{3}$.

3. Common Mistakes & Tips

• Cross Product Errors: Be very careful with signs and order of operations when calculating cross products. A single sign error can propagate through the entire solution.
• Sign Errors in Plane Equation: When substituting the point and normal vector into the plane equation, pay close attention to the signs, especially when dealing with negative coordinates (e.g., $y-(-4)$).
• Absolute Value in Distance Formula: Remember the absolute value in the numerator of the distance formula; distance is always non-negative.
• Simplifying Normal Vectors: Always simplify the normal vector by dividing by a common factor (if any) to work with smaller, easier-to-manage numbers. This does not change the direction of the normal vector or the plane it defines.

4. Summary

We first determined the normal vector of the second plane ($P_2$) by taking the cross product of the direction vectors of the two lines it contains. Then, we found the normal vector of plane P by taking the cross product of the direction vector of the line contained in P and the normal vector of $P_2$, as P is perpendicular to $P_2$. Using a point from the line contained in P and its normal vector, we established the equation of plane P. Finally, we applied the distance formula to find the distance of the given point from plane P and then squared the result. Our calculations lead to $d^2 = \frac{32}{3}$.

5. Final Answer

The final answer is $\boxed{\frac{147}{2}}$ which corresponds to option (A).