Question:

Step 1: Identify the given line and a general point on it. The equation of the given line is $\dfrac{x-0}{1}=\dfrac{y-3}{1}=\dfrac{z-1}{-1}$. Let this common ratio be $t$. Then, a general point $M$ on the line can be expressed as: $$ M(t, t+3, 1-t) $$ Step 2: Find the coordinates of the foot of the perpendicular from point $Q$ to the line. Let $M(t, t+3, 1-t)$ be the foot of the perpendicular from the point $Q(3,-3,1)$ to the line. The vector $\vec{QM}$ connects $Q$ to $M$: $$ \vec{QM} = (t-3, (t+3)-(-3), (1-t)-1) = (t-3, t+6, -t) $$ The direction vector of the line is $\vec{d} = (1,1,-1)$. Since $\vec{QM}$ is perpendicular to the line, their dot product is zero: $$ \vec{QM} \cdot \vec{d} = 0 $$ $$ (t-3)(1) + (t+6)(1) + (-t)(-1) = 0 $$ $$ t-3+t+6+t = 0 $$ $$ 3t+3 = 0 \Rightarrow t = -1 $$ Substitute $t=-1$ into the coordinates of $M$: $$ M(-1, -1+3, 1-(-1)) = M(-1, 2, 2) $$ Step 3: Determine the coordinates of the image point $P(\alpha,\beta,\gamma)$. Since $P$ is the image of $Q$ in the line, the foot of the perpendicular $M$ is the midpoint of the segment $QP$. Let $P(\alpha, \beta, \gamma)$. Using the midpoint formula: $$ M = \left(\frac{3+\alpha}{2}, \frac{-3+\beta}{2}, \frac{1+\gamma}{2}\right) $$ Equating the coordinates of $M$: $$ \frac{3+\alpha}{2} = -1 \Rightarrow 3+\alpha = -2 \Rightarrow \alpha = -5 $$ $$ \frac{-3+\beta}{2} = 2 \Rightarrow -3+\beta = 4 \Rightarrow \beta = 7 $$ $$ \frac{1+\gamma}{2} = 2 \Rightarrow 1+\gamma = 4 \Rightarrow \gamma = 3 $$ So, the image point is $P(-5, 7, 3)$. Step 4: Calculate the vectors $\vec{QP}$ and $\vec{QR}$. We have the points $Q(3,-3,1)$, $P(-5,7,3)$, and $R(2,5,-1)$. Calculate the vectors $\vec{QP}$ and $\vec{QR}$: $$ \vec{QP} = P - Q = (-5-3, 7-(-3), 3-1) = (-8, 10, 2) $$ $$ \vec{QR} = R - Q = (2-3, 5-(-3), -1-1) = (-1, 8, -2) $$ Step 5: Compute the cross product $\vec{QP} \times \vec{QR}$. The cross product of $\vec{QP}$ and $\vec{QR}$ is: $$ \vec{QP} \times \vec{QR} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ -8 & 10 & 2 \\ -1 & 8 & -2 \end{vmatrix} $$ $$ = \mathbf{i}((10)(-2) - (2)(8)) - \mathbf{j}((-8)(-2) - (2)(-1)) + \mathbf{k}((-8)(8) - (10)(-1)) $$ $$ = \mathbf{i}(-20 - 16) - \mathbf{j}(16 + 2) + \mathbf{k}(-64 + 10) $$ $$ = -36\mathbf{i} - 18\mathbf{j} - 54\mathbf{k} = (-36, -18, -54) $$ Step 6: Calculate the magnitude of the cross product to find the area of triangle $PQR$. The area of triangle $PQR$, denoted by $\lambda$, is half the magnitude of the cross product $\vec{QP} \times \vec{QR}$: $$ \lambda = \frac{1}{2} |\vec{QP} \times \vec{QR}| $$ $$ |\vec{QP} \times \vec{QR}| = \sqrt{(-36)^2 + (-18)^2 + (-54)^2} $$ We can factor out $18$ from each term: $$ |\vec{QP} \times \vec{QR}| = \sqrt{(18 \times -2)^2 + (18 \times -1)^2 + (18 \times -3)^2} $$ $$ = \sqrt{18^2((-2)^2 + (-1)^2 + (-3)^2)} $$ $$ = \sqrt{18^2(4 + 1 + 9)} = \sqrt{18^2 \times 14} = 18\sqrt{14} $$ Now, calculate the area $\lambda$: $$ \lambda = \frac{1}{2} (18\sqrt{14}) = 9\sqrt{14} $$ Step 7: Use the given relation $\lambda^2 = 14K$ to find the value of $K$. We have $\lambda = 9\sqrt{14}$. Square both sides: $$ \lambda^2 = (9\sqrt{14})^2 = 9^2 \times (\sqrt{14})^2 = 81 \times 14 $$ Given that $\lambda^2 = 14K$: $$ 81 \times 14 = 14K $$ Dividing both sides by 14: $$ K = 81 $$ Step 8: State the final answer. The value of $K$ is $81$. The correct option is Option 2.