Key Concepts and Formulas

1. Equation of a plane passing through the line of intersection of two given 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 two planes, then any plane passing through their line of intersection can be represented by the equation $P_1 + \lambda P_2 = 0$, where $\lambda$ is a scalar parameter. This represents a family of planes, and a specific value of $\lambda$ determines a unique plane satisfying an additional condition.
2. Condition for a plane to be perpendicular to a coordinate plane: A plane $Ax + By + Cz + D = 0$ is perpendicular to the xy-plane (whose equation is $z=0$, and normal vector is $\vec{n}_{xy} = \hat{k}$) if its normal vector, $\vec{n} = A\hat{i} + B\hat{j} + C\hat{k}$, is perpendicular to $\vec{n}_{xy}$. This implies $\vec{n} \cdot \vec{n}_{xy} = 0$, which means $C = 0$. Geometrically, this means the normal vector of the plane lies in the xy-plane, and thus the plane itself is perpendicular to the xy-plane.
3. Distance of a point from a plane: The perpendicular distance 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: Formulating the General Equation of Plane P

We are given two planes: $P_1: x + y + z – 6 = 0$ $P_2: 2x + 3y + z + 5 = 0$

Plane P contains the line of intersection of $P_1$ and $P_2$. According to the first key concept, the general equation of such a plane is given by $P_1 + \lambda P_2 = 0$.

Substituting the equations of $P_1$ and $P_2$: $(x + y + z – 6) + \lambda (2x + 3y + z + 5) = 0$

To express this in the standard form $Ax + By + Cz + D = 0$, we group the terms with $x$, $y$, $z$, and the constant terms: $x(1 + 2\lambda) + y(1 + 3\lambda) + z(1 + \lambda) + (-6 + 5\lambda) = 0$ This is the equation of plane P in terms of the parameter $\lambda$. Its normal vector is $\vec{n}_P = (1+2\lambda)\hat{i} + (1+3\lambda)\hat{j} + (1+\lambda)\hat{k}$.

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

We are given that plane P is perpendicular to the xy-plane. The normal vector of the xy-plane is $\vec{n}_{xy} = \hat{k}$.

For two planes to be perpendicular, their normal vectors must be perpendicular. This means their dot product must be zero: $\vec{n}_P \cdot \vec{n}_{xy} = 0$ $((1 + 2\lambda)\hat{i} + (1 + 3\lambda)\hat{j} + (1 + \lambda)\hat{k}) \cdot \hat{k} = 0$ Since $\hat{i} \cdot \hat{k} = 0$, $\hat{j} \cdot \hat{k} = 0$, and $\hat{k} \cdot \hat{k} = 1$, the dot product simplifies to: $(1 + \lambda) = 0$ Solving for $\lambda$: $\lambda = -1$ Reasoning: The coefficient of $z$ in the plane's equation represents the z-component of its normal vector. For the plane to be perpendicular to the xy-plane, its normal vector must lie entirely within the xy-plane, meaning its z-component must be zero. Setting $1+\lambda = 0$ achieves this.

Step 3: Determining the Specific Equation of Plane P

Now that we have the value of $\lambda = -1$, we substitute it back into the general equation of plane P from Step 1: $(x + y + z – 6) + (-1)(2x + 3y + z + 5) = 0$ $x + y + z – 6 - 2x - 3y - z - 5 = 0$

Combine like terms: $(1 - 2)x + (1 - 3)y + (1 - 1)z + (-6 - 5) = 0$ $-x - 2y + 0z - 11 = 0$ $-x - 2y - 11 = 0$

Multiplying the entire equation by -1 to make the leading coefficient positive (which is standard practice for clarity): $x + 2y + 11 = 0$ This is the specific equation of plane P. Note that the $z$ term is absent, which is consistent with the plane being perpendicular to the xy-plane.

Step 4: Calculating the Distance of the Point (0, 0, 256) from Plane P

We need to find the distance of the point $(x_0, y_0, z_0) = (0, 0, 256)$ from the plane $P: x + 2y + 0z + 11 = 0$. Here, the coefficients of the plane equation are $A=1$, $B=2$, $C=0$, and the constant term is $D=11$.

Using the distance formula: $D = \frac{|Ax_0 + By_0 + Cz_0 + D|}{\sqrt{A^2 + B^2 + C^2}}$ $D = \frac{|(1)(0) + (2)(0) + (0)(256) + 11|}{\sqrt{1^2 + 2^2 + 0^2}}$ $D = \frac{|0 + 0 + 0 + 11|}{\sqrt{1 + 4 + 0}}$ $D = \frac{|11|}{\sqrt{5}}$ $D = \frac{11}{\sqrt{5}}$

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Common Mistakes & Tips

• Sign Errors: Be meticulous with signs when expanding and combining terms, especially when $\lambda$ is negative.
• Misinterpreting Perpendicularity: Remember that "perpendicular to the xy-plane" means the coefficient of $z$ in the plane's equation must be zero, not necessarily that the plane equation itself is $z=0$.
• Irrelevant Coordinates: In this specific case, since plane P is perpendicular to the xy-plane (meaning its equation has no $z$ term), the z-coordinate of the point (256) does not influence the calculated distance.
• Absolute Value: Always include the absolute value in the numerator of the distance formula, as distance is a non-negative quantity.

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Summary

This problem required us to first determine the equation of a specific plane P by using the family of planes passing through the intersection of two given planes and then applying a perpendicularity condition. We found the parameter $\lambda = -1$, which led to the plane equation $x + 2y + 11 = 0$. Finally, we used the formula for the distance from a point to a plane to calculate the distance of $(0, 0, 256)$ from this plane. The calculated distance is $\frac{11}{\sqrt{5}}$.

The final answer is $\boxed{11/\sqrt{5}}$, which corresponds to option (C).