1. Key Concepts and Formulas

To solve this problem, we'll use fundamental concepts from 3D Geometry:

• Image of a point in a plane: Given a point $P_1(x_1, y_1, z_1)$ and a plane with equation $ax + by + cz + d = 0$, its image $P'(x', y', z')$ in the plane is found using the formula: $\frac{x' - x_1}{a} = \frac{y' - y_1}{b} = \frac{z' - z_1}{c} = \frac{-2(ax_1 + by_1 + cz_1 + d)}{a^2 + b^2 + c^2}$ Why this formula? The line segment connecting a point to its image is perpendicular to the plane. The direction ratios $(a, b, c)$ of the plane's normal vector are the same as the direction ratios of the line connecting the point and its image. The midpoint of the segment $P_1P'$ lies on the plane. The factor of $-2$ in the numerator reflects the point across the plane, placing the image twice the perpendicular distance from the point to the plane, on the opposite side.

• Area of a triangle using vectors: If $\vec{A}$ and $\vec{B}$ are two vectors representing two adjacent sides of a triangle originating from a common vertex, the area of the triangle is given by: $\text{Area} = \frac{1}{2} |\vec{A} \times \vec{B}|$ Why this formula? The magnitude of the cross product of two vectors, $|\vec{A} \times \vec{B}|$, is equal to the area of the parallelogram formed by these two vectors. A triangle formed by these two vectors shares the same base and height as half of this parallelogram, so its area is half the magnitude of their cross product.

2. Step-by-Step Solution

Our objective is to determine the area of $\triangle PQR$. We are provided with the coordinates of points $Q$ and $R$, but the coordinates of point $P$ are unknown. We will first find $P$ by using the given information about its image, and then proceed to calculate the triangle's area using vector methods.

Step 1: Determining the Coordinates of Point P

We are given that $Q(0, -1, -3)$ is the image of point $P$ in the plane $3x - y + 4z = 2$. This implies that $P$ is the original point $(x_1, y_1, z_1)$ and $Q$ is its image $(x', y', z')$. Alternatively, we can use the property that if $Q$ is the image of $P$, then $P$ is also the image of $Q$ in the same plane. This allows us to treat $Q$ as the initial point $(x_1, y_1, z_1)$ and find $P$ as its image $(x', y', z')$.

1. Identify the given information:

   • Point $Q(x_1, y_1, z_1) = (0, -1, -3)$.
   • Plane equation: $3x - y + 4z = 2$, which can be rewritten as $3x - y + 4z - 2 = 0$.
   • Comparing with $ax + by + cz + d = 0$, we have $a=3$, $b=-1$, $c=4$, and $d=-2$.
   • Let $P$ be $(x_P, y_P, z_P)$.

2. Apply the image formula: Substitute the values into the formula for the image of a point: $\frac{x_P - x_1}{a} = \frac{y_P - y_1}{b} = \frac{z_P - z_1}{c} = \frac{-2(ax_1 + by_1 + cz_1 + d)}{a^2 + b^2 + c^2}$ $\frac{x_P - 0}{3} = \frac{y_P - (-1)}{-1} = \frac{z_P - (-3)}{4} = \frac{-2(3(0) + (-1)(-1) + 4(-3) - 2)}{3^2 + (-1)^2 + 4^2}$

3. Calculate the common ratio: First, evaluate the numerator and denominator of the rightmost term:

   • Numerator: $-2(0 + 1 - 12 - 2) = -2(-13) = 26$
   • Denominator: $3^2 + (-1)^2 + 4^2 = 9 + 1 + 16 = 26$ So, the common ratio is $\frac{26}{26} = 1$.

   The equation system becomes: $\frac{x_P}{3} = \frac{y_P + 1}{-1} = \frac{z_P + 3}{4} = 1$

4. Solve for the coordinates of P:

   • From $\frac{x_P}{3} = 1 \Rightarrow x_P = 3$
   • From $\frac{y_P + 1}{-1} = 1 \Rightarrow y_P + 1 = -1 \Rightarrow y_P = -2$
   • From $\frac{z_P + 3}{4} = 1 \Rightarrow z_P + 3 = 4 \Rightarrow z_P = 1$

   Thus, the coordinates of point $P$ are $(3, -2, 1)$.

Step 2: Identifying the Vertices of $\triangle PQR$

Now that we have successfully found point $P$, we have all three vertices of the triangle:

• $P = (3, -2, 1)$
• $Q = (0, -1, -3)$
• $R = (3, -1, -2)$

Step 3: Calculating Vectors Representing Two Adjacent Sides of the Triangle

To apply the cross product formula for the area, we need two vectors originating from a common vertex. Let's choose vertex $Q$ as the common origin and form vectors $\vec{QP}$ and $\vec{QR}$.

1. Vector $\vec{QP}$ (from Q to P): This vector is obtained by subtracting the coordinates of $Q$ from $P$. $\vec{QP} = P - Q = (3 - 0)\hat{i} + (-2 - (-1))\hat{j} + (1 - (-3))\hat{k}$ $\vec{QP} = 3\hat{i} - \hat{j} + 4\hat{k}$

2. Vector $\vec{QR}$ (from Q to R): This vector is obtained by subtracting the coordinates of $Q$ from $R$. $\vec{QR} = R - Q = (3 - 0)\hat{i} + (-1 - (-1))\hat{j} + (-2 - (-3))\hat{k}$ $\vec{QR} = 3\hat{i} + 0\hat{j} + 1\hat{k}$

Step 4: Calculating the Cross Product of the Vectors

Next, we compute the cross product $\vec{QP} \times \vec{QR}$. The resulting vector will be perpendicular to the plane containing $\triangle PQR$. $\vec{QP} \times \vec{QR} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & -1 & 4 \\ 3 & 0 & 1 \end{vmatrix}$ Expand the determinant: $= \hat{i}((-1)(1) - (4)(0)) - \hat{j}((3)(1) - (4)(3)) + \hat{k}((3)(0) - (-1)(3))$ $= \hat{i}(-1 - 0) - \hat{j}(3 - 12) + \hat{k}(0 - (-3))$ $= -1\hat{i} - (-9)\hat{j} + 3\hat{k}$ $\vec{QP} \times \vec{QR} = -\hat{i} + 9\hat{j} + 3\hat{k}$

Step 5: Finding the Magnitude of the Cross Product

The magnitude of the cross product vector $\vec{QP} \times \vec{QR} = -\hat{i} + 9\hat{j} + 3\hat{k}$ is calculated using the formula $|\vec{V}| = \sqrt{V_x^2 + V_y^2 + V_z^2}$: $|\vec{QP} \times \vec{QR}| = \sqrt{(-1)^2 + (9)^2 + (3)^2}$ $= \sqrt{1 + 81 + 9}$ $= \sqrt{91}$

Step 6: Calculating the Area of Triangle PQR

Finally, the area of $\triangle PQR$ is half the magnitude of the cross product: $\text{Area}(\triangle PQR) = \frac{1}{2} |\vec{QP} \times \vec{QR}|$ $\text{Area}(\triangle PQR) = \frac{1}{2} \sqrt{91}$

3. Common Mistakes & Tips

• Image vs. Foot of Perpendicular: Remember that the formula for the image of a point uses a factor of $\mathbf{-2}$ in the numerator, while the formula for the foot of the perpendicular uses $\mathbf{-1}$. Double-check this factor to avoid errors.
• Vector Subtraction Order: When creating vectors like $\vec{QP}$, ensure you subtract the coordinates of the starting point ($Q$) from the endpoint ($P$), i.e., $P - Q$. The order affects the vector's direction and thus the cross product's direction (though not its magnitude).
• Cross Product Calculation: Be meticulous when calculating the determinant for the cross product. A single sign error or arithmetic mistake can lead to an incorrect final vector and magnitude.
• Forgetting the Half Factor: Always divide the magnitude of the cross product by 2 to get the area of the triangle. The cross product magnitude gives the area of the parallelogram formed by the vectors, not the triangle.

4. Summary

This problem effectively combines two crucial concepts in 3D geometry: finding the image of a point across a plane and calculating the area of a triangle using vector cross products. By accurately applying the image formula, we first determined the coordinates of the unknown vertex P. Subsequently, we formed two vectors representing adjacent sides of the triangle, calculated their cross product, and then found its magnitude. Dividing this magnitude by two yielded the area of $\triangle PQR$. This methodical approach ensures accuracy in multi-step 3D geometry problems.

The final answer is $\boxed{\frac{\sqrt{91}}{2}}$, which corresponds to option (C).