Key Concepts and Formulas

• Image of a Point in a Line: The image of a point $\mathrm{Q}$ in a line $L$ is a point $\mathrm{P}$ such that the line $L$ is the perpendicular bisector of the segment $\mathrm{QP}$. This means:
  • The line segment connecting $\mathrm{Q}$ and $\mathrm{P}$ is perpendicular to $L$.
  • The foot of the perpendicular from $\mathrm{Q}$ to $L$ (let's call it $\mathrm{M}$) is the midpoint of $\mathrm{QP}$.
• Vector Perpendicularity: Two vectors are perpendicular if and only if their dot product is zero. If $\vec{a}$ is perpendicular to $\vec{b}$, then $\vec{a} \cdot \vec{b} = 0$.
• Area of a Triangle in 3D: The area of a triangle with vertices $\mathrm{P}$, $\mathrm{Q}$, $\mathrm{R}$ can be calculated as half the magnitude of the cross product of two vectors representing two sides of the triangle originating from a common vertex. For example, $\text{Area} = \frac{1}{2} |\vec{PQ} \times \vec{PR}|$.

---

Step-by-Step Solution

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

Our first task is to find the coordinates of $\mathrm{P}(\alpha, \beta, \gamma)$, which is the image of $\mathrm{Q}(3,-3,1)$ in the given line $L: \frac{x-0}{1}=\frac{y-3}{1}=\frac{z-1}{-1}$.

1.1 Parameterize the Line and Define a General Point M A general point $\mathrm{M}$ on the line $L$ can be expressed in terms of a parameter, say $\mu$. Let $\frac{x}{1}=\frac{y-3}{1}=\frac{z-1}{-1} = \mu$. From this, we get the coordinates of $\mathrm{M}$: $x = \mu$ $y = \mu+3$ $z = -\mu+1$ So, $\mathrm{M}(\mu, \mu+3, -\mu+1)$ represents any point on the line $L$.

1.2 Apply the Perpendicularity Condition to find the Foot of the Perpendicular The line segment $\mathrm{QM}$ is perpendicular to the line $L$. First, determine the direction vector of the line $L$, which is given by the denominators in its symmetric form: $\vec{d_L} = (1, 1, -1)$. Next, form the vector $\vec{QM}$ connecting $\mathrm{Q}(3,-3,1)$ to $\mathrm{M}(\mu, \mu+3, -\mu+1)$: $\vec{QM} = (\mu-3, (\mu+3)-(-3), (-\mu+1)-1)$ $\vec{QM} = (\mu-3, \mu+6, -\mu)$ Since $\vec{QM}$ is perpendicular to $\vec{d_L}$, their dot product must be zero: $\vec{QM} \cdot \vec{d_L} = 0$ $(\mu-3)(1) + (\mu+6)(1) + (-\mu)(-1) = 0$ $\mu-3 + \mu+6 + \mu = 0$ $3\mu + 3 = 0$ $3\mu = -3$ $\mu = -1$ This value of $\mu$ corresponds to the specific point $\mathrm{M}$ on the line that is the foot of the perpendicular from $\mathrm{Q}$.

1.3 Find the Coordinates of the Foot of the Perpendicular M Substitute $\mu = -1$ back into the coordinates of $\mathrm{M}$: $\mathrm{M}(-1, (-1)+3, -(-1)+1)$ $\mathrm{M}(-1, 2, 2)$

1.4 Use the Midpoint Formula to Find P Since $\mathrm{P}$ is the image of $\mathrm{Q}$ in the line $L$, the foot of the perpendicular $\mathrm{M}$ is the midpoint of the line segment $\mathrm{PQ}$. Let $\mathrm{P} = (\alpha, \beta, \gamma)$. Using the midpoint formula: $\mathrm{M} = \left(\frac{x_Q+x_P}{2}, \frac{y_Q+y_P}{2}, \frac{z_Q+z_P}{2}\right)$ Substitute the coordinates of $\mathrm{M}(-1, 2, 2)$ and $\mathrm{Q}(3, -3, 1)$: For the x-coordinate: $-1 = \frac{3+\alpha}{2} \implies -2 = 3+\alpha \implies \alpha = -5$ For the y-coordinate: $2 = \frac{-3+\beta}{2} \implies 4 = -3+\beta \implies \beta = 7$ For the z-coordinate: $2 = \frac{1+\gamma}{2} \implies 4 = 1+\gamma \implies \gamma = 3$ Thus, the coordinates of point $\mathrm{P}$ are $(-5, 7, 3)$.

Step 2: Calculating the Area of Triangle PQR

Now we have all three vertices: $\mathrm{P}(-5, 7, 3)$, $\mathrm{Q}(3, -3, 1)$, and $\mathrm{R}(2, 5, -1)$. We will calculate the area of $\triangle \mathrm{PQR}$, denoted by $\lambda$, using the vector cross product method.

2.1 Calculate Vectors Representing Two Sides of the Triangle Let's choose $\vec{PQ}$ and $\vec{QR}$ as the two vectors originating from vertex $\mathrm{Q}$. $\vec{PQ} = \mathrm{Q} - \mathrm{P} = (3 - (-5), -3 - 7, 1 - 3)$ $\vec{PQ} = (8, -10, -2)$ $\vec{QR} = \mathrm{R} - \mathrm{Q} = (2 - 3, 5 - (-3), -1 - 1)$ $\vec{QR} = (-1, 8, -2)$

2.2 Compute the Cross Product $\vec{PQ} \times \vec{QR}$ The cross product is calculated as a determinant: $\vec{PQ} \times \vec{QR} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 8 & -10 & -2 \\ -1 & 8 & -2 \end{vmatrix}$ $= \hat{i}((-10)(-2) - (-2)(8)) - \hat{j}((8)(-2) - (-2)(-1)) + \hat{k}((8)(8) - (-10)(-1))$ $= \hat{i}(20 - (-16)) - \hat{j}(-16 - 2) + \hat{k}(64 - 10)$ $= \hat{i}(20 + 16) - \hat{j}(-18) + \hat{k}(54)$ $= 36\hat{i} + 18\hat{j} + 54\hat{k}$

2.3 Find the Magnitude of the Cross Product and the Area $\lambda$ The magnitude of the cross product is: $|\vec{PQ} \times \vec{QR}| = \sqrt{(36)^2 + (18)^2 + (54)^2}$ To simplify calculations, observe that $36 = 2 \times 18$, $18 = 1 \times 18$, and $54 = 3 \times 18$. $|\vec{PQ} \times \vec{QR}| = \sqrt{(2 \times 18)^2 + (1 \times 18)^2 + (3 \times 18)^2}$ $|\vec{PQ} \times \vec{QR}| = \sqrt{18^2 (2^2 + 1^2 + 3^2)}$ $|\vec{PQ} \times \vec{QR}| = \sqrt{18^2 (4 + 1 + 9)}$ $|\vec{PQ} \times \vec{QR}| = \sqrt{18^2 (14)}$ $|\vec{PQ} \times \vec{QR}| = 18\sqrt{14}$ The area of $\triangle \mathrm{PQR}$ is $\lambda = \frac{1}{2} |\vec{PQ} \times \vec{QR}|$: $\lambda = \frac{1}{2} (18\sqrt{14})$ $\lambda = 9\sqrt{14}$

Step 3: Determining the Value of K

We are given the relation $\lambda^2 = 14\mathrm{~K}$. We have found $\lambda = 9\sqrt{14}$. Let's calculate $\lambda^2$: $\lambda^2 = (9\sqrt{14})^2 = 9^2 \times (\sqrt{14})^2 = 81 \times 14$ Now, equate this to $14\mathrm{~K}$: $81 \times 14 = 14\mathrm{~K}$ Divide both sides by 14: $\mathrm{K} = 81$

---

Common Mistakes & Tips

• Sign Errors in Vector Operations: Be meticulous with positive and negative signs when calculating vector components, especially in the cross product. A single sign error can drastically alter the result.
• Order of Cross Product: While $|\vec{PQ} \times \vec{QR}| = |\vec{QR} \times \vec{PQ}|$, ensure consistent order within the determinant expansion to avoid errors. The magnitude will be the same, but the vector direction will be opposite.
• Simplifying Radicals: When calculating magnitudes, factoring out common terms (like $18^2$ in this case) before squaring and summing can significantly simplify arithmetic and prevent calculation errors with large numbers.
• Conceptual Link: Always remember why you are using a particular formula (e.g., dot product for perpendicularity, midpoint for image). This reinforces understanding and helps debug your steps.

---

Summary

The problem required us to first find the image of a point in a line, which was achieved by finding the foot of the perpendicular from the point to the line (using the dot product condition for orthogonality) and then using the midpoint formula. Once the coordinates of the image point were determined, we calculated the area of the triangle formed by the three points using the magnitude of the cross product of two side vectors. Finally, we used the given relation between the area squared and K to solve for K. All steps were executed carefully, leading to the value of K.

The final answer is $\boxed{81}$, which corresponds to option (B).