Key Concepts and Formulas

1. Equation of a Plane Containing Two Intersecting Lines: If two lines, $L_1$ passing through $(x_1, y_1, z_1)$ with direction vector $\vec{d_1} = \langle l_1, m_1, n_1 \rangle$, and $L_2$ passing through $(x_2, y_2, z_2)$ with direction vector $\vec{d_2} = \langle l_2, m_2, n_2 \rangle$, intersect, a unique plane contains them. The equation of this plane can be found using the determinant: $\begin{vmatrix} x - x_1 & y - y_1 & z - z_1 \\ l_1 & m_1 & n_1 \\ l_2 & m_2 & n_2 \end{vmatrix} = 0$ Here, $(x_1, y_1, z_1)$ can be any point on the plane (e.g., a point from $L_1$ or $L_2$). The normal vector to the plane is $\vec{n} = \vec{d_1} \times \vec{d_2}$.

2. Perpendicular Distance from a Point to a Plane: The perpendicular distance $D$ from a point $P(x_0, y_0, z_0)$ to a plane $Ax + By + Cz + D_{plane} = 0$ is given by the formula: $D = \frac{|Ax_0 + By_0 + Cz_0 + D_{plane}|}{\sqrt{A^2 + B^2 + C^2}}$

Step-by-Step Solution

Step 1: Determine the Equation of the Plane

The problem asks for the foot of the perpendicular from point P onto the plane containing two given lines. First, we need to find the equation of this plane. The two lines are given in symmetric (Cartesian) form: Line 1 ($L_1$): $\frac{x + 1}{6} = \frac{y - 1}{7} = \frac{z - 3}{8}$ Line 2 ($L_2$): $\frac{x - 1}{3} = \frac{y - 2}{5} = \frac{z - 3}{7}$

1.1 Extract Points and Direction Vectors From $L_1$:

• A point on the line is $A(-1, 1, 3)$.
• Its direction vector is $\vec{d_1} = \langle 6, 7, 8 \rangle$.

From $L_2$:

• A point on the line is $B(1, 2, 3)$.
• Its direction vector is $\vec{d_2} = \langle 3, 5, 7 \rangle$.

1.2 Verify if the Lines Intersect For a unique plane to contain two lines, they must either intersect or be parallel.

• Check for Parallelism: We compare the direction vectors $\vec{d_1} = \langle 6, 7, 8 \rangle$ and $\vec{d_2} = \langle 3, 5, 7 \rangle$. Since their components are not proportional (e.g., $6/3 = 2$, but $7/5 \neq 2$), the lines are not parallel.

• Check for Intersection: We set the parametric forms of the lines equal to each other. $L_1: x = -1 + 6\lambda, y = 1 + 7\lambda, z = 3 + 8\lambda$ $L_2: x = 1 + 3\mu, y = 2 + 5\mu, z = 3 + 7\mu$

  Equating the $z$-coordinates: $3 + 8\lambda = 3 + 7\mu \Rightarrow 8\lambda = 7\mu \quad \mathbf{(1)}$

  Equating the $x$-coordinates: $-1 + 6\lambda = 1 + 3\mu \Rightarrow 6\lambda - 3\mu = 2 \quad \mathbf{(2)}$

  Substitute $\mu = \frac{8}{7}\lambda$ from (1) into (2): $6\lambda - 3\left(\frac{8}{7}\lambda\right) = 2$ $6\lambda - \frac{24}{7}\lambda = 2$ $\frac{42\lambda - 24\lambda}{7} = 2$ $\frac{18\lambda}{7} = 2 \Rightarrow \lambda = \frac{14}{18} = \frac{7}{9}$

  Now we find $\mu$ using $\lambda = \frac{7}{9}$: $\mu = \frac{8}{7} \cdot \frac{7}{9} = \frac{8}{9}$

  Finally, we check if the $y$-coordinates match for these values of $\lambda$ and $\mu$: For $L_1: y = 1 + 7\left(\frac{7}{9}\right) = 1 + \frac{49}{9} = \frac{9+49}{9} = \frac{58}{9}$ For $L_2: y = 2 + 5\left(\frac{8}{9}\right) = 2 + \frac{40}{9} = \frac{18+40}{9} = \frac{58}{9}$

  Since all coordinates match for specific values of $\lambda$ and $\mu$, the lines intersect. This confirms that a unique plane contains both lines.

1.3 Apply the Determinant Formula for the Plane Equation We use point $A(-1, 1, 3)$ from $L_1$ as $(x_1, y_1, z_1)$ and the direction vectors $\vec{d_1} = \langle 6, 7, 8 \rangle$ and $\vec{d_2} = \langle 3, 5, 7 \rangle$. The equation of the plane is: $\begin{vmatrix} x - (-1) & y - 1 & z - 3 \\ 6 & 7 & 8 \\ 3 & 5 & 7 \end{vmatrix} = 0$ $\begin{vmatrix} x + 1 & y - 1 & z - 3 \\ 6 & 7 & 8 \\ 3 & 5 & 7 \end{vmatrix} = 0$

1.4 Expand the Determinant and Simplify Expanding the determinant along the first row: $(x+1)(7 \times 7 - 8 \times 5) - (y-1)(6 \times 7 - 8 \times 3) + (z-3)(6 \times 5 - 7 \times 3) = 0$ $(x+1)(49 - 40) - (y-1)(42 - 24) + (z-3)(30 - 21) = 0$ $(x+1)(9) - (y-1)(18) + (z-3)(9) = 0$

To simplify, we divide the entire equation by 9: $(x+1) - 2(y-1) + (z-3) = 0$ $x + 1 - 2y + 2 + z - 3 = 0$ $x - 2y + z = 0$ This is the equation of the plane containing the two given lines.

Step 2: Calculate the Perpendicular Distance (PQ) from Point P to the Plane

We need to find the perpendicular distance from point $P(7, -2, 13)$ to the plane $x - 2y + z = 0$.

2.1 Identify Point Coordinates and Plane Coefficients

• Point coordinates: $(x_0, y_0, z_0) = (7, -2, 13)$.
• Plane equation: $x - 2y + z = 0$. Comparing this with the general plane equation $Ax + By + Cz + D_{plane} = 0$, we have: $A = 1$ $B = -2$ $C = 1$ $D_{plane} = 0$

2.2 Apply the Distance Formula Substitute these values into the perpendicular distance formula: $PQ = \frac{|A x_0 + B y_0 + C z_0 + D_{plane}|}{\sqrt{A^2 + B^2 + C^2}}$ $PQ = \frac{|1(7) + (-2)(-2) + 1(13) + 0|}{\sqrt{1^2 + (-2)^2 + 1^2}}$ $PQ = \frac{|7 + 4 + 13|}{\sqrt{1 + 4 + 1}}$ $PQ = \frac{|\sqrt{6}|}{\sqrt{6}}$ $PQ = 1$

Step 3: Calculate $(PQ)^2$

The problem asks for the square of the distance, $(PQ)^2$: $(PQ)^2 = (1)^2$ $(PQ)^2 = 1$

Common Mistakes & Tips

• Verifying Line Properties: Always check if the lines are parallel or intersecting. If they are skew, a single plane cannot contain both lines. If they are parallel, the plane equation is found differently.
• Sign Errors: Be meticulous with signs when substituting coordinates into the determinant and the distance formula. A small sign error can lead to an incorrect result.
• Determinant Expansion: Double-check the calculation of minors and cofactors when expanding the determinant.
• Simplifying the Plane Equation: After finding the plane equation, dividing by a common factor (if any) simplifies the coefficients, making subsequent distance calculations easier.
• Final Step: Ensure you calculate exactly what the question asks for (e.g., distance, square of distance, foot of perpendicular, etc.). In this case, it was $(PQ)^2$.

Summary

To find $(PQ)^2$:

1. We first determined the equation of the plane containing the two given lines. We confirmed that the lines intersect, allowing us to use the determinant method. The plane equation was found to be $x - 2y + z = 0$.
2. Next, we calculated the perpendicular distance $PQ$ from the point $P(7, -2, 13)$ to this plane using the standard distance formula. The distance $PQ$ was calculated as $1$.
3. Finally, we squared this distance to find $(PQ)^2$, which resulted in $1$.

The final answer is $\boxed{1}$.