Consider the lines $$L_1 : x - 1 = y - 2 = z$$ and $$L_2 : x - 2 = y = z - 1$$. Let the feet of the perpendiculars from the point $$P(5, 1, -3)$$ on the lines $$L_1$$ and $$L_2$$ be Q and R respectively. If the area of the triangle PQR is A, then $$4A^2$$ is equal to :

The symmetric form of $$L_1$$ is $$x-1=y-2=z=\lambda$$, so a convenient point on $$L_1$$ is $$A_1(1,2,0)$$ and the direction vector is $$\mathbf{v}=\langle 1,1,1\rangle$$.

For a point $$P(x_0,y_0,z_0)$$, the foot of the perpendicular $$Q$$ on a line through $$A_1$$ with direction $$\mathbf{v}$$ is given by
$$Q=A_1+\dfrac{(P-A_1)\cdot\mathbf{v}}{|\mathbf{v}|^{2}}\;\mathbf{v}$$ $$-(1)$$.
Here $$P(5,1,-3)$$ and $$A_1(1,2,0)$$, so $$P-A_1=\langle 4,-1,-3\rangle$$ and$$(P-A_1)\cdot\mathbf{v}=4+(-1)+(-3)=0.$$
Using $$(1)$$ with numerator $$0$$ gives $$Q=A_1=(1,2,0).$$ Thus $$Q(1,2,0).$$

The symmetric form of $$L_2$$ is $$x-2=y=z-1=\mu$$, so a point on $$L_2$$ is $$A_2(2,0,1)$$ with the same direction $$\mathbf{v}=\langle 1,1,1\rangle$$.

Applying $$-(1)$$ to $$L_2$$: $$P-A_2=\langle 3,1,-4\rangle$$ and$$(P-A_2)\cdot\mathbf{v}=3+1+(-4)=0.$$
Hence $$R=A_2=(2,0,1).$$ Thus $$R(2,0,1).$$

Compute the vectors of the sides of $$\triangle PQR$$:
$$\overrightarrow{PQ}=Q-P=\langle -4,1,3\rangle,$$
$$\overrightarrow{PR}=R-P=\langle -3,-1,4\rangle.$$

The area of a triangle whose sides are $$\mathbf{a}$$ and $$\mathbf{b}$$ is $$\dfrac{1}{2}\,|\mathbf{a}\times\mathbf{b}|$$. First find the cross product:
$$\overrightarrow{PQ}\times\overrightarrow{PR}=\begin{vmatrix} \mathbf{i}&\mathbf{j}&\mathbf{k}\\ -4&\;1&\;3\\ -3&-1&\;4\end{vmatrix}$$
$$=\mathbf{i}(1\cdot4-3\cdot(-1))-\mathbf{j}((-4)\cdot4-3\cdot(-3))+\mathbf{k}((-4)\cdot(-1)-1\cdot(-3))$$
$$=\mathbf{i}(4+3)-\mathbf{j}(-16+9)+\mathbf{k}(4+3)$$
$$=\langle 7,7,7\rangle.$$

Magnitude of the cross product:$$|\langle 7,7,7\rangle|=\sqrt{7^{2}+7^{2}+7^{2}}=\sqrt{147}=7\sqrt{3}.$$

Therefore, the area of $$\triangle PQR$$ is
$$A=\dfrac{1}{2}\,(7\sqrt{3})=\dfrac{7}{2}\sqrt{3}.$$

Finally, $$4A^{2}=4\left(\dfrac{7}{2}\sqrt{3}\right)^{2}=4\left(\dfrac{49\cdot3}{4}\right)=147.$$

Hence $$4A^{2}=147$$, which corresponds to Option B.