Let A be the point of intersection of the lines $$L_1 : \frac{x-7}{1} = \frac{y-5}{0} = \frac{z-3}{-1}$$ and $$L_2 : \frac{x-1}{3} = \frac{y+3}{4} = \frac{z+7}{5}$$. Let B and C be the points on the lines $$L_1$$ and $$L_2$$ respectively such that $$AB = AC = \sqrt{15}$$. Then the square of the area of the triangle ABC is :

Write the two lines in parametric form.

For $$L_1$$, put parameter $$t$$:
$$\frac{x-7}{1}=\frac{y-5}{0}=\frac{z-3}{-1}\; \Longrightarrow \; x=7+t,\; y=5,\; z=3-t$$

For $$L_2$$, put parameter $$s$$:
$$\frac{x-1}{3}=\frac{y+3}{4}=\frac{z+7}{5}\; \Longrightarrow \; x=1+3s,\; y=-3+4s,\; z=-7+5s$$

To find the point of intersection A, equate the coordinates:

$$7+t = 1+3s,\quad 5 = -3+4s,\quad 3-t = -7+5s$$

From $$5 = -3+4s$$ we get $$4s=8$$, so $$s=2$$.

Substituting $$s=2$$ in the $$x$$-equation gives $$7+t = 1+6$$, hence $$t=0$$.
The $$z$$-equation is also satisfied. Therefore

$$A(7,\,5,\,3).$$

Next locate point $$B$$ on $$L_1$$. Let its parameter be $$t=k$$:

$$B(7+k,\,5,\,3-k).$$

Distance formula gives
$$AB^2 = (k)^2 + 0^2 + (-k)^2 = 2k^2.$$ Given $$AB = \sqrt{15}$$, so

$$2k^2 = 15 \;\Longrightarrow\; k^2 = \frac{15}{2}. \;-(1)$$

Similarly, point $$C$$ lies on $$L_2$$. Write $$s = 2 + \ell$$:

$$C(7+3\ell,\,5+4\ell,\,3+5\ell).$$

Distance AC:
$$AC^2 = (3\ell)^2 + (4\ell)^2 + (5\ell)^2 = 9\ell^2 +16\ell^2 +25\ell^2 = 50\ell^2.$$ Given $$AC = \sqrt{15}$$, so

$$50\ell^2 = 15 \;\Longrightarrow\; \ell^2 = \frac{3}{10}. \;-(2)$$

Take any sign choices for $$k$$ and $$\ell$$; the area depends on $$k^2$$ and $$\ell^2$$ only.

Find vectors $$\overrightarrow{AB}$$ and $$\overrightarrow{AC}$$:

$$\overrightarrow{AB} = (k,\,0,\,-k),\quad \overrightarrow{AC} = (3\ell,\,4\ell,\,5\ell).$$

The cross product is

$$\overrightarrow{AB}\times\overrightarrow{AC} = \begin{vmatrix} \mathbf{i}&\mathbf{j}&\mathbf{k}\\[2pt] k&0&-k\\[2pt] 3\ell&4\ell&5\ell \end{vmatrix} = (4k\ell)\,\mathbf{i} \;-\; (8k\ell)\,\mathbf{j} \;+\; (4k\ell)\,\mathbf{k}.$$ Thus
$$\lvert\overrightarrow{AB}\times\overrightarrow{AC}\rvert^2 = (4k\ell)^2(1^2+(-2)^2+1^2) = 16k^2\ell^2 \times 6 = 96k^2\ell^2.$$

Area of triangle ABC is
$$\frac{1}{2}\lvert\overrightarrow{AB}\times\overrightarrow{AC}\rvert,$$ so the square of the area equals

$$\left(\frac{1}{2}\right)^2 \times 96k^2\ell^2 = 24k^2\ell^2.$$ Insert $$k^2$$ from $$(1)$$ and $$\ell^2$$ from $$(2)$$:

$$24 \times \frac{15}{2} \times \frac{3}{10} = 24 \times \frac{45}{20} = 24 \times \frac{9}{4} = 6 \times 9 = 54.$$

Therefore, the square of the area of $$\triangle ABC$$ is $$54$$.
Hence Option A is correct.