Key Concepts and Formulas

1. Equation of a Plane Passing Through the Intersection of Two Planes: If we have two planes, $P_1: A_1x+B_1y+C_1z+D_1=0$ and $P_2: A_2x+B_2y+C_2z+D_2=0$, then the equation of any plane passing through their line of intersection is given by $P_1 + \lambda P_2 = 0$, where $\lambda$ is a scalar constant. This equation represents a "family of planes."
2. Condition for Perpendicular Planes: Two planes are perpendicular if and only if their normal vectors are perpendicular. The normal vector to a plane $Ax+By+Cz+D=0$ is $\vec{n} = A\hat{i} + B\hat{j} + C\hat{k}$. Thus, for perpendicular planes with normal vectors $\vec{n}_A = A_A\hat{i} + B_A\hat{j} + C_A\hat{k}$ and $\vec{n}_B = A_B\hat{i} + B_B\hat{j} + C_B\hat{k}$, their dot product must be zero: $\vec{n}_A \cdot \vec{n}_B = 0$, which means $A_AA_B + B_AB_B + C_AC_B = 0$.
3. Distance of a Point from a Plane: The distance $d$ of a point $(x_0, y_0, z_0)$ from a plane $Ax+By+Cz+D=0$ is given by the formula: $d = \frac{|Ax_0+By_0+Cz_0+D|}{\sqrt{A^2+B^2+C^2}}$

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Step-by-Step Solution

Step 1: Formulate the General Equation of Plane P

We are given two planes whose intersection forms a line through which plane P passes. To use the formula for a plane passing through the intersection of two planes, we first rewrite the given plane equations in the standard form $Ax+By+Cz+D=0$: Plane 1 ($P_1$): $2x+3y-z=2 \implies 2x+3y-z-2=0$ Plane 2 ($P_2$): $x+2y+3z=6 \implies x+2y+3z-6=0$

The equation of any plane passing through the intersection of $P_1$ and $P_2$ is given by $P_1 + \lambda P_2 = 0$. Substituting the equations: $(2x+3y-z-2) + \lambda(x+2y+3z-6) = 0$ To identify the normal vector of this plane, we group the terms by $x, y, z$: $x(2+\lambda) + y(3+2\lambda) + z(-1+3\lambda) + (-2-6\lambda) = 0$ This is the general equation of plane P. Its normal vector, $\vec{n}_P$, is $(2+\lambda)\hat{i} + (3+2\lambda)\hat{j} + (-1+3\lambda)\hat{k}$.

Step 2: Apply the Perpendicularity Condition to find $\lambda$

We are given that plane P is perpendicular to the plane $P_3: 2x+y-z+1=0$. The normal vector of plane $P_3$ is $\vec{n}_3 = 2\hat{i} + 1\hat{j} - 1\hat{k}$.

Since plane P is perpendicular to plane $P_3$, their normal vectors must be perpendicular. This means their dot product must be zero: $\vec{n}_P \cdot \vec{n}_3 = 0$. $(2+\lambda)(2) + (3+2\lambda)(1) + (-1+3\lambda)(-1) = 0$ Now, we solve this equation for $\lambda$: $4+2\lambda + 3+2\lambda + 1-3\lambda = 0$ Combine the constant terms and the $\lambda$ terms: $(4+3+1) + (2\lambda+2\lambda-3\lambda) = 0$ $8 + \lambda = 0$ $\lambda = -8$

Step 3: Determine the Specific Equation of Plane P

Now that we have the value of $\lambda = -8$, we substitute it back into the general equation of plane P from Step 1: $x(2+(-8)) + y(3+2(-8)) + z(-1+3(-8)) + (-2-6(-8)) = 0$ $x(2-8) + y(3-16) + z(-1-24) + (-2+48) = 0$ $-6x - 13y - 25z + 46 = 0$ For convenience, we can multiply the entire equation by -1 to make the leading coefficient positive: $6x + 13y + 25z - 46 = 0$ This is the specific equation of plane P. From this, we identify the coefficients $A=6$, $B=13$, $C=25$, and $D=-46$.

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

We need to find the distance $d$ of plane P ($6x + 13y + 25z - 46 = 0$) from the point $(-7, 1, 1)$. Using the distance formula for a point $(x_0, y_0, z_0)$ from a plane $Ax+By+Cz+D=0$: $d = \frac{|Ax_0+By_0+Cz_0+D|}{\sqrt{A^2+B^2+C^2}}$ Here, $(x_0, y_0, z_0) = (-7, 1, 1)$, and $A=6, B=13, C=25, D=-46$. $d = \frac{|(6)(-7) + (13)(1) + (25)(1) + (-46)|}{\sqrt{6^2 + 13^2 + 25^2}}$ $d = \frac{|-42 + 13 + 25 - 46|}{\sqrt{36 + 169 + 625}}$ $d = \frac{|-42 + 38 - 46|}{\sqrt{830}}$ $d = \frac{|-4 - 46|}{\sqrt{830}}$ $d = \frac{|-50|}{\sqrt{830}}$ $d = \frac{50}{\sqrt{830}}$

Step 5: Calculate $d^2$

The question asks for $d^2$. We square the distance $d$ we just calculated: $d^2 = \left(\frac{50}{\sqrt{830}}\right)^2$ $d^2 = \frac{50^2}{(\sqrt{830})^2}$ $d^2 = \frac{2500}{830}$ To simplify the fraction, we can divide both the numerator and the denominator by 10: $d^2 = \frac{250}{83}$

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Common Mistakes & Tips

• Sign Errors: Be extremely careful with negative signs, especially when multiplying terms in the dot product or substituting coordinates into the distance formula. A single sign error can lead to an incorrect value of $\lambda$ or an incorrect distance.
• Absolute Value: Remember to use the absolute value in the numerator of the distance formula. Distance is always a non-negative quantity.
• Normal Vectors: Ensure you correctly identify the normal vector coefficients ($A, B, C$) from the plane equation.

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Summary

This problem required us to find the equation of a plane P satisfying two conditions: passing through the intersection of two given planes and being perpendicular to a third plane. We first formed a general equation for plane P using the parameter $\lambda$. Then, we utilized the perpendicularity condition of normal vectors to solve for $\lambda$. Once $\lambda$ was determined, we found the specific equation of plane P. Finally, we applied the formula for the distance of a point from a plane to calculate $d$ and then $d^2$.

The final answer is $\boxed{\frac{250}{83}}$, which corresponds to option (A).