Key Concepts and Formulas

1. Equation of a Line in 3D: A line passing through a point $(x_0, y_0, z_0)$ with direction ratios $(a, b, c)$ can be represented in symmetric form as $\frac{x - x_0}{a} = \frac{y - y_0}{b} = \frac{z - z_0}{c}$. Any point on this line can be expressed parametrically as $(x_0 + ak, y_0 + bk, z_0 + ck)$ for some scalar $k$.
2. Foot of the Perpendicular from a Point to a Line: If M is the foot of the perpendicular from a point $P_0$ to a line L, then the vector $P_0M$ is perpendicular to the direction vector of L. Their dot product must be zero.
3. Mirror Image of a Point with respect to a Line: If M is the foot of the perpendicular from point $P_0$ to line L, and Q is the mirror image of $P_0$ with respect to L, then M is the midpoint of the line segment $P_0Q$. That is, $M = \left( \frac{x_{P_0} + x_Q}{2}, \frac{y_{P_0} + y_Q}{2}, \frac{z_{P_0} + z_Q}{2} \right)$.
4. Equation of a Plane: A plane passing through a point $(x_1, y_1, z_1)$ with a normal vector $\vec{n} = (A, B, C)$ has the equation $A(x - x_1) + B(y - y_1) + C(z - z_1) = 0$. If a line L is perpendicular to a plane P, then the direction vector of L serves as the normal vector for P.

Step-by-Step Solution

Step 1: Identify the characteristics of the given line L. The equation of line L is given by $\frac{x - 3}{2} = \frac{y - 1}{1} = \frac{z - 2}{1}$. From this symmetric form, we can identify:

• A point on the line L: $A = (3, 1, 2)$.
• The direction vector of the line L: $\vec{d} = (2, 1, 1)$. Any point M on the line L can be represented parametrically as $M = (3 + 2k, 1 + k, 2 + k)$ for some scalar $k$.

Step 2: Find the foot of the perpendicular (M) from point (2, 3, -1) to line L. Let $P_0 = (2, 3, -1)$. Let M be the foot of the perpendicular from $P_0$ to L. The vector $P_0M$ connects $P_0$ to a point M on L. $P_0M = ((3 + 2k) - 2, (1 + k) - 3, (2 + k) - (-1))$ $P_0M = (1 + 2k, k - 2, k + 3)$. Since $P_0M$ is perpendicular to line L, its dot product with the direction vector $\vec{d}$ must be zero: $P_0M \cdot \vec{d} = 0$ $(1 + 2k)(2) + (k - 2)(1) + (k + 3)(1) = 0$ $2 + 4k + k - 2 + k + 3 = 0$ $6k + 3 = 0$ $6k = -3$ $k = -\frac{1}{2}$. Now, substitute the value of $k$ back into the parametric coordinates of M to find the coordinates of the foot of the perpendicular: $x_M = 3 + 2(-\frac{1}{2}) = 3 - 1 = 2$ $y_M = 1 + (-\frac{1}{2}) = \frac{1}{2}$ $z_M = 2 + (-\frac{1}{2}) = \frac{3}{2}$ So, the foot of the perpendicular is $M = (2, \frac{1}{2}, \frac{3}{2})$.

Step 3: Find the mirror image (Q) of point (2, 3, -1) with respect to line L. Since M is the foot of the perpendicular and Q is the mirror image of $P_0$ with respect to L, M is the midpoint of the segment $P_0Q$. Let $Q = (x_Q, y_Q, z_Q)$. Using the midpoint formula: $x_M = \frac{x_{P_0} + x_Q}{2} \Rightarrow 2 = \frac{2 + x_Q}{2} \Rightarrow 4 = 2 + x_Q \Rightarrow x_Q = 2$ $y_M = \frac{y_{P_0} + y_Q}{2} \Rightarrow \frac{1}{2} = \frac{3 + y_Q}{2} \Rightarrow 1 = 3 + y_Q \Rightarrow y_Q = -2$ $z_M = \frac{z_{P_0} + z_Q}{2} \Rightarrow \frac{3}{2} = \frac{-1 + z_Q}{2} \Rightarrow 3 = -1 + z_Q \Rightarrow z_Q = 4$ Thus, the mirror image point is $Q = (2, -2, 4)$.

Step 4: Determine the equation of plane P. The plane P passes through Q, and line L is perpendicular to P. This means the direction vector of L, $\vec{d} = (2, 1, 1)$, is the normal vector to the plane P. So, $\vec{n} = (2, 1, 1)$. The plane P passes through $Q = (2, -2, 4)$. Using the equation of a plane $A(x - x_1) + B(y - y_1) + C(z - z_1) = 0$: $2(x - 2) + 1(y - (-2)) + 1(z - 4) = 0$ $2(x - 2) + (y + 2) + (z - 4) = 0$ $2x - 4 + y + 2 + z - 4 = 0$ $2x + y + z - 6 = 0$. This is the equation of plane P.

Step 5: Check which of the given points lies on the plane P. Substitute the coordinates of each option into the plane equation $2x + y + z - 6 = 0$: (A) $(-1, 1, 2)$: $2(-1) + 1 + 2 - 6 = -2 + 1 + 2 - 6 = -5 \neq 0$. (B) $(1, 1, 1)$: $2(1) + 1 + 1 - 6 = 4 - 6 = -2 \neq 0$. (C) $(1, 1, 2)$: $2(1) + 1 + 2 - 6 = 5 - 6 = -1 \neq 0$. (D) $(1, 2, 2)$: $2(1) + 2 + 2 - 6 = 2 + 2 + 2 - 6 = 6 - 6 = 0$. Point (1, 2, 2) satisfies the equation of plane P, meaning it lies on the plane.

Common Mistakes & Tips

• Sign Errors: Be extremely careful with negative signs, especially when substituting coordinates or performing dot products. A single sign error can lead to a completely different result.
• Parametric Form: Always convert the line equation to its parametric form to easily represent any point on the line. This is crucial for finding the foot of the perpendicular.
• Vector Perpendicularity: Remember that two vectors are perpendicular if and only if their dot product is zero. This is a fundamental concept for finding the foot of the perpendicular.
• Midpoint Formula: Ensure correct application of the midpoint formula for finding the mirror image. It's $Q = 2M - P_0$.
• Normal Vector: The normal vector of a plane perpendicular to a line is the direction vector of that line.

Summary

We first found the parametric form of the line L and used it to determine the foot of the perpendicular (M) from the given point (2, 3, -1) to the line. With M as the midpoint, we calculated the coordinates of the mirror image point (Q). Finally, using Q and the direction vector of L (which serves as the normal vector for the plane P), we derived the equation of plane P. Checking the given options revealed which point lies on this plane. Based on our calculations, the plane is $2x+y+z-6=0$, and the point (1,2,2) lies on this plane.

The final answer is $\boxed{(-1, 1, 2)}$.