Let $$Q$$ be the mirror image of the point $$P(1, 0, 1)$$ with respect to the plane $$S : x + y + z = 5$$. If a line $$L$$ passing through $$(1, -1, -1)$$, parallel to the line $$PQ$$ meets the plane $$S$$ at $$R$$, then $$QR^2$$ is equal to

We have the point $$P(1, 0, 1)$$ and the plane $$S: x + y + z = 5$$. First, the normal vector of the plane is $$\vec{n} = (1, 1, 1)$$, which leads to the distance
$$ d = \frac{|1 + 0 + 1 - 5|}{\sqrt{3}} = \frac{3}{\sqrt{3}} = \sqrt{3}. $$

Consequently, the mirror image $$Q$$ of $$P$$ with respect to $$S$$ is given by
$$ Q = P + 2 \cdot \frac{5 - (1+0+1)}{3} \cdot (1,1,1) = (1,0,1) + 2 \cdot 1 \cdot (1,1,1) = (3, 2, 3). $$ Since the vector from $$P$$ to $$Q$$ is $$\vec{PQ} = (2, 2, 2)$$, the direction ratios are $$(1, 1, 1)$$.

Next, if one considers the line $$L$$ through $$(1, -1, -1)$$ parallel to $$(1, 1, 1)$$, it can be written as
$$ L: (1+t,\ -1+t,\ -1+t). $$ Substituting this parametrization into the plane equation produces
$$ (1+t) + (-1+t) + (-1+t) = 5 \\ 3t - 1 = 5 \implies t = 2, $$ and hence the intersection point is $$R = (3, 1, 1)$$.

Finally, the squared distance between $$Q$$ and $$R$$ becomes
$$ QR^2 = (3-3)^2 + (2-1)^2 + (3-1)^2 = 0 + 1 + 4 = 5. $$ Therefore, the answer is Option B: $$5$$.