Let  $$P$$  be the image of the point  $$Q(7,-2,5)$$ in the line  $$L:\;\frac{x-1}{2}=\frac{y+1}{3}=\frac{z}{4},$$ and  $$R(5,p,q)$$  be a point on $$L.$$ Then the square of the area of  $$\triangle PQR$$ is  $$\underline{\hspace{2cm}}.$$

Parametrizing $$L$$ as $$(1+2t, -1+3t, 4t)$$ and substituting $$1 + 2t = 5$$ gives $$t = 2$$, which yields $$p = -1 + 6 = 5$$ and $$q = 8$$, so that $$R = (5,5,8)$$.

To find the image $$P$$ of $$Q(7,-2,5)$$ in the line, let the foot of the perpendicular from $$Q$$ onto $$L$$ be $$M = (1+2t, -1+3t, 4t)$$ so that $$\vec{QM} = (2t-6, 3t+1, 4t-5)$$ is perpendicular to the direction vector $$(2,3,4)$$ of $$L$$. Imposing $$2(2t-6) + 3(3t+1) + 4(4t-5) = 0$$ gives $$4t-12 + 9t+3 + 16t-20 = 0$$, hence $$29t - 29 = 0 \Rightarrow t = 1$$, and thus $$M = (3,2,4)$$.

Since $$P$$ is the reflection of $$Q$$ across the line, we have $$M = \frac{P+Q}{2}$$ and therefore $$P = 2M - Q = (6-7,4-(-2),8-5) = (-1,6,3)$$.

Next, we compute the area of triangle $$PQR$$ by forming the vectors $$\vec{QP} = P - Q = (-1-7,6-(-2),3-5) = (-8,8,-2)$$ and $$\vec{QR} = R - Q = (5-7,5-(-2),8-5) = (-2,7,3)$$.

Their cross product is given by $$\vec{QP} \times \vec{QR} = \begin{vmatrix} \vec{i} & \vec{j} & \vec{k} \\ -8 & 8 & -2 \\ -2 & 7 & 3 \end{vmatrix}$$ which equals $$= \vec{i}(24-(-14)) - \vec{j}(-24-4) + \vec{k}(-56-(-16))$$, simplifies to $$= \vec{i}(24+14) - \vec{j}(-28) + \vec{k}(-56+16)$$, and hence $$= 38\vec{i} + 28\vec{j} - 40\vec{k}$$. This gives $$|\vec{QP} \times \vec{QR}|^2 = 38^2 + 28^2 + 40^2 = 1444 + 784 + 1600 = 3828$$. 

Since the area is $$\frac{1}{2}|\vec{QP} \times \vec{QR}|$$, its square is $$(\text{Area})^2 = \frac{1}{4} \times 3828 = 957$$.