Key Concepts and Formulas

• Image of a Point with Respect to a Plane: The image $Q(x_2, y_2, z_2)$ of a point $P(x_1, y_1, z_1)$ with respect to a plane $ax+by+cz+d=0$ is found using the formula: $\frac{x_2 - x_1}{a} = \frac{y_2 - y_1}{b} = \frac{z_2 - z_1}{c} = \frac{-2(ax_1 + by_1 + cz_1 + d)}{a^2 + b^2 + c^2}$ This formula is based on the fact that the line segment PQ is perpendicular to the plane, and its midpoint lies on the plane.
• Equation of a Plane: The equation of a plane passing through a point $(x_0, y_0, z_0)$ with a normal vector $\vec{n} = a\widehat{i} + b\widehat{j} + c\widehat{k}$ is given by: $a(x-x_0) + b(y-y_0) + c(z-z_0) = 0$
• Normal Vector from Two Vectors in a Plane: If two non-parallel vectors $\vec{u}$ and $\vec{v}$ lie in a plane (or are parallel to the plane), their cross product $\vec{u} \times \vec{v}$ gives a vector perpendicular to both, and thus a normal vector to the plane.

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Step-by-Step Solution

Step 1: Finding the Coordinates of Point Q (the Image of P)

Our first objective is to find the mirror image of point $P(1, 2, 1)$ with respect to the plane $x + 2y + 2z = 16$. This image point is denoted as Q.

1. Identify parameters:

   • Point $P(x_1, y_1, z_1) = (1, 2, 1)$.
   • Plane equation: $x + 2y + 2z - 16 = 0$. Comparing this to $ax+by+cz+d=0$, we have $a=1, b=2, c=2, d=-16$.

2. Apply the image formula: Let $Q(x_2, y_2, z_2)$ be the image. Substitute the values into the formula: $\frac{x_2 - 1}{1} = \frac{y_2 - 2}{2} = \frac{z_2 - 1}{2} = \frac{-2(1(1) + 2(2) + 2(1) - 16)}{1^2 + 2^2 + 2^2}$

3. Calculate the right-hand side:

   • Numerator: $ax_1 + by_1 + cz_1 + d = 1(1) + 2(2) + 2(1) - 16 = 1 + 4 + 2 - 16 = 7 - 16 = -9$.
   • So, $-2(ax_1 + by_1 + cz_1 + d) = -2(-9) = 18$.
   • Denominator: $a^2 + b^2 + c^2 = 1^2 + 2^2 + 2^2 = 1 + 4 + 4 = 9$.
   • Therefore, the common ratio is $\frac{18}{9} = 2$.

4. Solve for $x_2, y_2, z_2$:

   • $\frac{x_2 - 1}{1} = 2 \implies x_2 - 1 = 2 \implies x_2 = 3$
   • $\frac{y_2 - 2}{2} = 2 \implies y_2 - 2 = 4 \implies y_2 = 6$
   • $\frac{z_2 - 1}{2} = 2 \implies z_2 - 1 = 4 \implies z_2 = 5$

   Thus, the coordinates of the mirror image point are $Q(3, 6, 5)$.

Step 2: Extracting Information for Plane T

We are given that plane T passes through point Q and contains the line $\vec{r} = - \widehat k + \lambda \left( {\widehat i + \widehat j + 2\widehat k} \right)$. To define a plane, we need a point on the plane and its normal vector.

1. Point on Plane T: We already have $Q(3, 6, 5)$.
2. Information from the line: The equation of the line $\vec{r} = \vec{a} + \lambda \vec{d}$ tells us:
   • A point on the line (and thus on plane T): $\vec{a} = -\widehat{k}$, so $A(0, 0, -1)$.
   • A direction vector parallel to the line (and thus parallel to plane T): $\vec{d} = \widehat{i} + \widehat{j} + 2\widehat{k}$.

Now we have two distinct points on plane T, $Q(3, 6, 5)$ and $A(0, 0, -1)$, and a vector $\vec{d}$ that is parallel to the plane.

Step 3: Determining the Normal Vector of Plane T

Since plane T contains the line and passes through Q, any vector formed by points on the plane, or the direction vector of the line, must lie within or be parallel to the plane. The normal vector to the plane will be perpendicular to any such vectors.

1. Identify two vectors parallel to the plane:

   • The direction vector of the line: $\vec{d} = \widehat{i} + \widehat{j} + 2\widehat{k}$.
   • A vector connecting the two points on the plane, A and Q: $\vec{AQ} = Q - A = (3-0)\widehat{i} + (6-0)\widehat{j} + (5-(-1))\widehat{k} = 3\widehat{i} + 6\widehat{j} + 6\widehat{k}$. For simpler calculation, we can use a scalar multiple of $\vec{AQ}$, such as $\vec{AQ'} = \widehat{i} + 2\widehat{j} + 2\widehat{k}$ (by dividing $\vec{AQ}$ by 3). This vector is also parallel to plane T.

2. Calculate the normal vector $\vec{n}$ using the cross product: The normal vector $\vec{n}$ is found by taking the cross product of $\vec{d}$ and $\vec{AQ'}$: $\vec{n} = \vec{AQ'} \times \vec{d} = (\widehat{i} + 2\widehat{j} + 2\widehat{k}) \times (\widehat{i} + \widehat{j} + 2\widehat{k})$ Using the determinant form: $\vec{n} = \begin{vmatrix} \widehat{i} & \widehat{j} & \widehat{k} \\ 1 & 2 & 2 \\ 1 & 1 & 2 \end{vmatrix}$ $\vec{n} = \widehat{i}((2)(2) - (2)(1)) - \widehat{j}((1)(2) - (2)(1)) + \widehat{k}((1)(1) - (2)(1))$ $\vec{n} = \widehat{i}(4 - 2) - \widehat{j}(2 - 2) + \widehat{k}(1 - 2)$ $\vec{n} = 2\widehat{i} - 0\widehat{j} - \widehat{k} = 2\widehat{i} - \widehat{k}$ So, the normal vector to plane T is $\vec{n} = (2, 0, -1)$.

Step 4: Forming the Equation of Plane T

Now we have a normal vector $\vec{n} = (2, 0, -1)$ and a point on the plane, $A(0, 0, -1)$. We can use these to write the equation of plane T.

1. Apply the plane equation formula: Using $a=2, b=0, c=-1$ and $(x_0, y_0, z_0) = (0, 0, -1)$: $2(x - 0) + 0(y - 0) - 1(z - (-1)) = 0$ $2x + 0y - 1(z + 1) = 0$ $2x - z - 1 = 0$ Therefore, the equation of plane T is $2x - z = 1$.

Step 5: Verifying Which Option Lies on Plane T

To check which of the given options lies on plane T, we substitute the coordinates of each point into the plane's equation, $2x - z = 1$.

• (A) $(2, 1, 0)$: Substitute $x=2, y=1, z=0$: $2(2) - 0 = 4$. Since $4 \neq 1$, point (A) does not lie on T.
• (B) $(1, 2, 1)$: Substitute $x=1, y=2, z=1$: $2(1) - 1 = 2 - 1 = 1$. Since $1 = 1$, point (B) lies on T.
• (C) $(1, 2, 2)$: Substitute $x=1, y=2, z=2$: $2(1) - 2 = 2 - 2 = 0$. Since $0 \neq 1$, point (C) does not lie on T.
• (D) $(1, 3, 2)$: Substitute $x=1, y=3, z=2$: $2(1) - 2 = 2 - 2 = 0$. Since $0 \neq 1$, point (D) does not lie on T.

Only point (1, 2, 1) satisfies the equation of plane T.

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Common Mistakes & Tips

• Sign Errors in Image Formula: Pay close attention to the -2 factor and the sign of d when calculating ax_1 + by_1 + cz_1 + d. A common mistake is to forget the -2 or miscalculate the sign.
• Cross Product Calculation: Ensure accuracy in the determinant expansion for the cross product. A small arithmetic or sign error will lead to an incorrect normal vector and thus an incorrect plane equation.
• Choosing Points for Plane Equation: When forming the plane equation $a(x-x_0) + b(y-y_0) + c(z-z_0) = 0$, pick the point $(x_0, y_0, z_0)$ with simpler coordinates (e.g., more zeros) to minimize arithmetic during substitution.

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Summary

This problem required a sequential application of 3D geometry concepts. First, we determined the coordinates of point Q, the mirror image of P, using the specific formula for point reflection across a plane. This yielded $Q(3, 6, 5)$. Next, we used the information about plane T (passing through Q and containing a given line) to identify two points on the plane ($Q$ and a point $A(0, 0, -1)$ from the line) and a direction vector $\vec{d}$ parallel to the plane. We then found a normal vector to plane T by taking the cross product of $\vec{d}$ and the vector $\vec{AQ'}$, resulting in $\vec{n} = 2\widehat{i} - \widehat{k}$. Finally, we used this normal vector and point A to derive the equation of plane T as $2x - z = 1$. By substituting the coordinates of the given options into this equation, we identified that point (1, 2, 1) lies on plane T.

The final answer is $\boxed{(1, 2, 1)}$, which corresponds to option (B).