Key Concepts and Formulas

This problem requires the application of two fundamental concepts from 3D Geometry:

1. Equation of a Plane Passing Through the Line of Intersection of Two Planes: If $P_1: A_1x + B_1y + C_1z + D_1 = 0$ and $P_2: A_2x + B_2y + C_2z + D_2 = 0$ are the equations of two distinct planes, then the equation of any plane that passes through their line of intersection is given by: $P_1 + \lambda P_2 = 0$ where $\lambda$ is an arbitrary scalar constant (a parameter). This equation represents a family of planes, and a specific value of $\lambda$ defines a unique plane within this family.

2. Distance of a Point from a Plane: The perpendicular distance $d$ of a point $(x_0, y_0, z_0)$ from a plane with the Cartesian equation $Ax+By+Cz+D=0$ 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: Convert the Given Plane Equations to Cartesian Form

The problem provides the equations of two planes in vector form. To effectively use the $P_1 + \lambda P_2 = 0$ formula, it's convenient to convert them into their standard Cartesian form, $Ax+By+Cz+D=0$. Recall that the position vector $\vec{r}$ represents a general point $(x,y,z)$ on the plane, so $\vec{r} = x\hat{i} + y\hat{j} + z\hat{k}$.

• Plane 1 ($P_1$): $\vec{r} \cdot(\hat{i}+\hat{j}+\hat{k})=6$ Substitute $\vec{r} = x\hat{i} + y\hat{j} + z\hat{k}$: $(x\hat{i} + y\hat{j} + z\hat{k}) \cdot(\hat{i}+\hat{j}+\hat{k})=6$ Performing the dot product: $x(1) + y(1) + z(1) = 6$ $x+y+z=6$ To match the $Ax+By+Cz+D=0$ format, move the constant term to the left side: $P_1: x+y+z-6=0$

• Plane 2 ($P_2$): $\vec{r} \cdot(2 \hat{i}+3 \hat{j}+4 \hat{k})=-5$ Substitute $\vec{r} = x\hat{i} + y\hat{j} + z\hat{k}$: $(x\hat{i} + y\hat{j} + z\hat{k}) \cdot(2 \hat{i}+3 \hat{j}+4 \hat{k})=-5$ Performing the dot product: $x(2) + y(3) + z(4) = -5$ $2x+3y+4z=-5$ Moving the constant term to the left side: $P_2: 2x+3y+4z+5=0$

Step 2: Formulate the General Equation of Plane P

Plane P contains the line of intersection of $P_1$ and $P_2$. Using the key concept for planes passing through the line of intersection, its equation is given by $P_1 + \lambda P_2 = 0$. This general equation represents a family of planes, all sharing the same line of intersection.

Substitute the Cartesian forms of $P_1$ and $P_2$ found in Step 1: $(x+y+z-6) + \lambda(2x+3y+4z+5) = 0$ Now, group the terms by $x$, $y$, and $z$ to express the equation in the standard form $Ax+By+Cz+D=0$: $(1+2\lambda)x + (1+3\lambda)y + (1+4\lambda)z + (-6+5\lambda) = 0$ This is the general equation of plane P, where $\lambda$ is an unknown parameter that we need to determine.

Step 3: Determine the Value of $\lambda$

The problem provides an additional condition: plane P passes through the point $(0,2,-2)$. This condition is crucial because it allows us to identify the unique plane P from the family of planes represented by the general equation. We substitute the coordinates of the point $(0,2,-2)$ into the general equation of plane P:

$(1+2\lambda)(0) + (1+3\lambda)(2) + (1+4\lambda)(-2) + (-6+5\lambda) = 0$ Simplify and solve for $\lambda$: $0 + (2+6\lambda) + (-2-8\lambda) + (-6+5\lambda) = 0$ Combine the constant terms and the $\lambda$ terms: $(2 - 2 - 6) + (6\lambda - 8\lambda + 5\lambda) = 0$ $-6 + 3\lambda = 0$ $3\lambda = 6$ $\lambda = 2$ This value of $\lambda=2$ uniquely defines the plane P that satisfies all given conditions.

Step 4: Find the Specific Equation of Plane P

Now that we have determined $\lambda=2$, we substitute this value back into the general equation of plane P (from Step 2) to find its specific equation: $(1+2(2))x + (1+3(2))y + (1+4(2))z + (-6+5(2)) = 0$ $(1+4)x + (1+6)y + (1+8)z + (-6+10) = 0$ $5x + 7y + 9z + 4 = 0$ This is the final Cartesian equation of plane P.

Step 5: Calculate the Distance of the Point $(12,12,18)$ from Plane P

The problem asks for the square of the distance of the point $(12,12,18)$ from plane P. First, we calculate the perpendicular distance $d$ using the distance formula. The point is $(x_0, y_0, z_0) = (12,12,18)$ and the plane is $5x+7y+9z+4=0$. Comparing with $Ax+By+Cz+D=0$, we have $A=5$, $B=7$, $C=9$, and $D=4$.

Substitute these values into the distance formula: $d = \frac{|Ax_0+By_0+Cz_0+D|}{\sqrt{A^2+B^2+C^2}}$ $d = \frac{|5(12) + 7(12) + 9(18) + 4|}{\sqrt{5^2+7^2+9^2}}$ Calculate the numerator: $5(12) = 60$ $7(12) = 84$ $9(18) = 162$ $\text{Numerator} = |60 + 84 + 162 + 4| = |310| = 310$ Calculate the denominator: $5^2 = 25$ $7^2 = 49$ $9^2 = 81$ $\text{Denominator} = \sqrt{25+49+81} = \sqrt{155}$ So, the distance $d$ is: $d = \frac{310}{\sqrt{155}}$

Step 6: Calculate the Square of the Distance

The question specifically asks for the square of the distance, $d^2$: $d^2 = \left(\frac{310}{\sqrt{155}}\right)^2$ $d^2 = \frac{310^2}{155}$ $d^2 = \frac{310 \times 310}{155}$ Notice that $310$ is exactly $2 \times 155$. $d^2 = \frac{(2 \times 155) \times 310}{155}$ $d^2 = 2 \times 310$ $d^2 = 620$

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Common Mistakes & Tips

• Sign Convention for Constant Term: When setting up $P_1 + \lambda P_2 = 0$, ensure both $P_1$ and $P_2$ are in the form $Ax+By+Cz+D=0$. For example, $x+y+z=6$ must be rewritten as $x+y+z-6=0$, not $x+y+z+6=0$. Incorrectly handling the sign of the constant term $D$ is a common error.
• Arithmetic Precision: Double-check all arithmetic calculations, especially when solving for $\lambda$ and substituting values into the distance formula. A small error can lead to a completely incorrect final answer.
• Understanding Vector Form: Remember that $\vec{r} \cdot \vec{n} = d$ corresponds to the Cartesian equation $Ax+By+Cz = d$, where $\vec{n} = A\hat{i}+B\hat{j}+C\hat{k}$. To use it in $P_1 + \lambda P_2 = 0$, it must be $Ax+By+Cz-d=0$.

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Summary

This problem effectively tests the understanding of a plane passing through the intersection of two planes and the distance of a point from a plane. The solution involved converting the given vector equations of planes into Cartesian form, using the $P_1 + \lambda P_2 = 0$ family of planes, and then determining the unique value of $\lambda$ by utilizing the condition that the plane passes through a specific point. Finally, the equation of plane P was used to calculate the distance of another point from it, and then the square of this distance was found.

The final answer is $\boxed{620}$, which corresponds to option (B).