Key Concepts and Formulas

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

1. Equation of a Plane Equidistant from Two Points: A plane whose points are equidistant from two given points $A(x_1, y_1, z_1)$ and $B(x_2, y_2, z_2)$ is the perpendicular bisector plane of the line segment $AB$. Its equation can be determined by setting the squared distance from a generic point $P(x, y, z)$ on the plane to $A$ equal to its squared distance to $B$, i.e., $PA^2 = PB^2$. The distance formula in 3D between two points $(x_a, y_a, z_a)$ and $(x_b, y_b, z_b)$ is $\sqrt{(x_b-x_a)^2 + (y_b-y_a)^2 + (z_b-z_a)^2}$.
2. Angle Between Two Planes: The acute angle $\theta$ between two planes $A_1x + B_1y + C_1z + D_1 = 0$ and $A_2x + B_2y + C_2z + D_2 = 0$ is found using their normal vectors $\vec{n_1} = (A_1, B_1, C_1)$ and $\vec{n_2} = (A_2, B_2, C_2)$. The formula is: $\cos \theta = \left| \frac{\vec{n_1} \cdot \vec{n_2}}{|\vec{n_1}| |\vec{n_2}|} \right|$ Here, $\vec{n_1} \cdot \vec{n_2}$ represents the dot product of the normal vectors, and $|\vec{n}|$ denotes the magnitude of a vector. The absolute value ensures that the calculated angle $\theta$ is acute.

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Step-by-Step Solution

Step 1: Determine the Equation of Plane P

• Understanding the geometric property: Plane P is defined as the locus of all points $(x, y, z)$ that are equidistant from the given points $A(-4, 2, 1)$ and $B(2, -2, 3)$. This means plane P is the perpendicular bisector of the line segment $AB$.
• Setting up the distance equality: Let $P(x, y, z)$ be any point on plane P. By definition, $PA = PB$. To simplify calculations and remove square roots, we equate the squared distances: $PA^2 = PB^2$
• Applying the squared distance formula: $(x - (-4))^2 + (y - 2)^2 + (z - 1)^2 = (x - 2)^2 + (y - (-2))^2 + (z - 3)^2$ $(x + 4)^2 + (y - 2)^2 + (z - 1)^2 = (x - 2)^2 + (y + 2)^2 + (z - 3)^2$
• Expanding the binomials: $(x^2 + 8x + 16) + (y^2 - 4y + 4) + (z^2 - 2z + 1) = (x^2 - 4x + 4) + (y^2 + 4y + 4) + (z^2 - 6z + 9)$
• Simplifying the equation: Notice that the $x^2, y^2, z^2$ terms cancel out from both sides. $8x - 4y - 2z + 21 = -4x + 4y - 6z + 17$
• Rearranging terms to the standard plane equation form ($Ax + By + Cz + D = 0$): $(8x + 4x) + (-4y - 4y) + (-2z + 6z) + (21 - 17) = 0$ $12x - 8y + 4z + 4 = 0$
• Simplifying by dividing by the common factor 4: $3x - 2y + z + 1 = 0$ This is the equation of plane P.
• Identifying the normal vector of Plane P: From the equation $3x - 2y + z + 1 = 0$, the normal vector to plane P is $\vec{n_1} = (3, -2, 1)$.

Step 2: Identify the Normal Vector of the Second Plane

• Given equation: The second plane is given by the equation $2x + y + 3z = 1$.
• Identifying the normal vector: From this equation, the normal vector to the second plane is $\vec{n_2} = (2, 1, 3)$.

Step 3: Calculate the Acute Angle Between the Two Planes

• Apply the formula for the angle between planes: The acute angle $\theta$ between plane P (with normal $\vec{n_1}$) and the second plane (with normal $\vec{n_2}$) is given by: $\cos \theta = \left| \frac{\vec{n_1} \cdot \vec{n_2}}{|\vec{n_1}| |\vec{n_2}|} \right|$
• Calculate the dot product $\vec{n_1} \cdot \vec{n_2}$: $\vec{n_1} \cdot \vec{n_2} = (3)(2) + (-2)(1) + (1)(3) = 6 - 2 + 3 = 7$
• Calculate the magnitude of $\vec{n_1}$: $|\vec{n_1}| = \sqrt{3^2 + (-2)^2 + 1^2} = \sqrt{9 + 4 + 1} = \sqrt{14}$
• Calculate the magnitude of $\vec{n_2}$: $|\vec{n_2}| = \sqrt{2^2 + 1^2 + 3^2} = \sqrt{4 + 1 + 9} = \sqrt{14}$
• Substitute these values into the cosine formula: $\cos \theta = \left| \frac{7}{\sqrt{14} \cdot \sqrt{14}} \right|$ $\cos \theta = \left| \frac{7}{14} \right|$ $\cos \theta = \frac{1}{2}$
• Determine the angle $\theta$: We need to find the angle whose cosine is $1/2$. $\theta = \arccos\left(\frac{1}{2}\right)$ $\theta = \frac{\pi}{3}$

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Common Mistakes & Tips

• Alternative Method for Plane Equation: Instead of using the distance formula, one can find the midpoint $M$ of the segment $AB$ and the vector $\vec{AB}$. The plane P passes through $M$ and its normal vector is parallel to $\vec{AB}$.
  • Midpoint $M = \left(\frac{-4+2}{2}, \frac{2-2}{2}, \frac{1+3}{2}\right) = (-1, 0, 2)$.
  • Vector $\vec{AB} = (2 - (-4), -2 - 2, 3 - 1) = (6, -4, 2)$.
  • The equation of the plane is $A(x-x_M) + B(y-y_M) + C(z-z_M) = 0$, where $(A, B, C)$ is the normal vector $(6, -4, 2)$.
  • $6(x - (-1)) - 4(y - 0) + 2(z - 2) = 0$
  • $6(x + 1) - 4y + 2z - 4 = 0$
  • $6x + 6 - 4y + 2z - 4 = 0$
  • $6x - 4y + 2z + 2 = 0$. Dividing by 2 gives $3x - 2y + z + 1 = 0$, which matches our result and is often a quicker method.
• Algebraic Precision: Be meticulous with signs and calculations when expanding squared terms and simplifying the plane equation. A small error can propagate and lead to an incorrect normal vector.
• Acute Angle Requirement: Always remember to use the absolute value in the $\cos \theta$ formula for the angle between planes to ensure you find the acute angle. If the absolute value is omitted, the result could be an obtuse angle.

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Summary

This problem involved two key steps: first, determining the equation of a plane that is the perpendicular bisector of a line segment connecting two given points, and second, calculating the acute angle between this newly found plane and another given plane. We found the equation of plane P by equating squared distances, yielding $3x - 2y + z + 1 = 0$. Then, we identified its normal vector $\vec{n_1} = (3, -2, 1)$ and the normal vector of the second plane $\vec{n_2} = (2, 1, 3)$. Finally, using the dot product formula for the angle between planes, we calculated $\cos \theta = 1/2$, which corresponds to an acute angle of $\frac{\pi}{3}$.

The final answer is $\boxed{\frac{\pi}{3}}$, which corresponds to option (C).