1. Key Concepts and Formulas

To determine the acute angle between the two planes, $P_1$ and $P_2$, we will use the following fundamental concepts from 3D Geometry:

• Family of Planes: If two planes are given by the equations $P_A: A_1x + B_1y + C_1z + D_1 = 0$ and $P_B: A_2x + B_2y + C_2z + D_2 = 0$, then any plane passing through their line of intersection can be represented by the general equation $P_A + \lambda P_B = 0$, where $\lambda$ is a scalar parameter. This expands to: $(A_1x + B_1y + C_1z + D_1) + \lambda (A_2x + B_2y + C_2z + D_2) = 0$ A specific value of $\lambda$ defines a unique plane in this family, usually determined by an additional condition, such as the plane passing through a given point.

• Angle Between Two Planes: The acute angle $\theta$ between two planes is found by calculating the angle between their respective normal vectors. If the normal vectors to the planes are $\vec{n_1} = \langle A_1, B_1, C_1 \rangle$ and $\vec{n_2} = \langle A_2, B_2, C_2 \rangle$, the cosine of the acute angle $\theta$ is given by the formula: $\cos \theta = \frac{|\vec{n_1} \cdot \vec{n_2}|}{|\vec{n_1}| |\vec{n_2}|}$ The absolute value in the numerator, $|\vec{n_1} \cdot \vec{n_2}|$, ensures that $\cos \theta \ge 0$, which corresponds to the acute angle (i.e., $\theta \in [0, \pi/2]$).

2. Step-by-Step Solution

Step 1: Formulate the General Equation for the Family of Planes

We are given two planes whose intersection defines the family: Plane $P_A: 5x + 8y + 13z - 29 = 0$ Plane $P_B: 8x - 7y + z - 20 = 0$

Any plane passing through the line of intersection of $P_A$ and $P_B$ can be expressed as $P_A + \lambda P_B = 0$. We write this equation and then group terms by $x, y, z$ to easily identify the normal vector components: $(5x + 8y + 13z - 29) + \lambda (8x - 7y + z - 20) = 0$ Rearranging the terms, we get the general equation for any plane in this family: $(5 + 8\lambda)x + (8 - 7\lambda)y + (13 + \lambda)z - (29 + 20\lambda) = 0 \quad (*)$ The normal vector to any plane in this family will be $\vec{n} = \langle 5 + 8\lambda, 8 - 7\lambda, 13 + \lambda \rangle$.

**Step 2: Determine Plane $P_1$ and its Normal Vector $\vec{n_1}$}

Plane $P_1$ is a member of the family $(*)$ and passes through the point $(2, 1, 3)$. To find the specific value of $\lambda$ (let's call it $\lambda_1$) for $P_1$, we substitute the coordinates $(x=2, y=1, z=3)$ into equation $(*)$: $(5 + 8\lambda_1)(2) + (8 - 7\lambda_1)(1) + (13 + \lambda_1)(3) - (29 + 20\lambda_1) = 0$ Now, we simplify and solve for $\lambda_1$: $(10 + 16\lambda_1) + (8 - 7\lambda_1) + (39 + 3\lambda_1) - (29 + 20\lambda_1) = 0$ Combine the $\lambda_1$ terms and the constant terms: $(16\lambda_1 - 7\lambda_1 + 3\lambda_1 - 20\lambda_1) + (10 + 8 + 39 - 29) = 0$ $(-8\lambda_1) + (28) = 0$ $8\lambda_1 = 28$ $\lambda_1 = \frac{28}{8} = \frac{7}{2}$ Now, we find the normal vector $\vec{n_1}$ for $P_1$ by substituting $\lambda_1 = \frac{7}{2}$ into the coefficients of $x, y, z$ from equation $(*)$:

• Coefficient of $x$: $A_1 = 5 + 8\left(\frac{7}{2}\right) = 5 + 28 = 33$
• Coefficient of $y$: $B_1 = 8 - 7\left(\frac{7}{2}\right) = 8 - \frac{49}{2} = \frac{16 - 49}{2} = -\frac{33}{2}$
• Coefficient of $z$: $C_1 = 13 + \left(\frac{7}{2}\right) = \frac{26 + 7}{2} = \frac{33}{2}$ So, $\vec{n_1} = \left\langle 33, -\frac{33}{2}, \frac{33}{2} \right\rangle$. For simplicity in calculations, we can use a scalar multiple of this vector. Multiplying by $\frac{2}{33}$ gives us a simpler normal vector: $\vec{n_1}' = \left\langle 33 \times \frac{2}{33}, -\frac{33}{2} \times \frac{2}{33}, \frac{33}{2} \times \frac{2}{33} \right\rangle = \langle 2, -1, 1 \rangle$

**Step 3: Determine Plane $P_2$ and its Normal Vector $\vec{n_2}$}

Plane $P_2$ also belongs to the family $(*)$ and passes through the point $(0, 1, 2)$. We substitute $(x=0, y=1, z=2)$ into equation $(*)$ to find $\lambda_2$: $(5 + 8\lambda_2)(0) + (8 - 7\lambda_2)(1) + (13 + \lambda_2)(2) - (29 + 20\lambda_2) = 0$ Simplify and solve for $\lambda_2$: $0 + (8 - 7\lambda_2) + (26 + 2\lambda_2) - (29 + 20\lambda_2) = 0$ Combine the $\lambda_2$ terms and the constant terms: $(-7\lambda_2 + 2\lambda_2 - 20\lambda_2) + (8 + 26 - 29) = 0$ $(-25\lambda_2) + (5) = 0$ $25\lambda_2 = 5$ $\lambda_2 = \frac{5}{25} = \frac{1}{5}$ Now, we find the normal vector $\vec{n_2}$ for $P_2$ by substituting $\lambda_2 = \frac{1}{5}$ into the coefficients of $x, y, z$ from equation $(*)$:

• Coefficient of $x$: $A_2 = 5 + 8\left(\frac{1}{5}\right) = 5 + \frac{8}{5} = \frac{25 + 8}{5} = \frac{33}{5}$
• Coefficient of $y$: $B_2 = 8 - 7\left(\frac{1}{5}\right) = 8 - \frac{7}{5} = \frac{40 - 7}{5} = \frac{33}{5}$
• Coefficient of $z$: $C_2 = 13 + \left(\frac{1}{5}\right) = \frac{65 + 1}{5} = \frac{66}{5}$ So, $\vec{n_2} = \left\langle \frac{33}{5}, \frac{33}{5}, \frac{66}{5} \right\rangle$. Again, we simplify by multiplying by $\frac{5}{33}$: $\vec{n_2}' = \left\langle \frac{33}{5} \times \frac{5}{33}, \frac{33}{5} \times \frac{5}{33}, \frac{66}{5} \times \frac{5}{33} \right\rangle = \langle 1, 1, 2 \rangle$

**Step 4: Calculate the Acute Angle Between $P_1$ and $P_2$}

We now have the simplified normal vectors for $P_1$ and $P_2$: $\vec{n_1}' = \langle 2, -1, 1 \rangle$ $\vec{n_2}' = \langle 1, 1, 2 \rangle$

We use the formula for the cosine of the acute angle $\theta$: $\cos \theta = \frac{|\vec{n_1}' \cdot \vec{n_2}'|}{|\vec{n_1}'| |\vec{n_2}'|}$ First, calculate the dot product: $\vec{n_1}' \cdot \vec{n_2}' = (2)(1) + (-1)(1) + (1)(2) = 2 - 1 + 2 = 3$ Next, calculate the magnitudes of the normal vectors: $|\vec{n_1}'| = \sqrt{2^2 + (-1)^2 + 1^2} = \sqrt{4 + 1 + 1} = \sqrt{6}$ $|\vec{n_2}'| = \sqrt{1^2 + 1^2 + 2^2} = \sqrt{1 + 1 + 4} = \sqrt{6}$ Substitute these values into the angle formula: $\cos \theta = \frac{|3|}{\sqrt{6} \times \sqrt{6}} = \frac{3}{6} = \frac{1}{2}$ Since $\cos \theta = \frac{1}{2}$, the acute angle $\theta$ is: $\theta = \arccos\left(\frac{1}{2}\right) = \frac{\pi}{3}$

3. Common Mistakes & Tips

• Sign Errors: Be very careful with signs when substituting coordinates into the general plane equation and when expanding and combining terms. A single sign error can lead to an incorrect $\lambda$ value and, subsequently, incorrect normal vectors.
• Forgetting Absolute Value: When calculating the acute angle between planes (or lines), remember to take the absolute value of the dot product in the numerator of the cosine formula. This ensures that $\cos \theta$ is positive, yielding an acute angle ($0 \le \theta \le \pi/2$).
• Simplifying Normal Vectors: Always simplify normal vectors by dividing out common factors. This makes calculations for dot products and magnitudes much easier and reduces the chance of arithmetic errors, without affecting the direction of the vector or the angle. For example, $\langle 33, -\frac{33}{2}, \frac{33}{2} \rangle$ is directionally equivalent to $\langle 2, -1, 1 \rangle$.

4. Summary

This problem required us to first determine the equations of two specific planes, $P_1$ and $P_2$, each belonging to a family of planes passing through the intersection of two given planes. This was achieved by using the family of planes formula $P_A + \lambda P_B = 0$ and then using the additional point condition for each plane to solve for its unique $\lambda$ value. Once the $\lambda$ values were found, we extracted the normal vectors for $P_1$ and $P_2$, simplifying them to make calculations easier. Finally, the acute angle between these two planes was calculated using the dot product formula involving their normal vectors. The result was $\frac{\pi}{3}$.

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