A straight line drawn from the point $$P(1,3,2)$$, parallel to the line $$\frac{x-2}{1} = \frac{y-4}{2} = \frac{z-6}{1}$$, intersects the plane $$L_1 : x - y + 3z = 6$$ at the point $$Q$$. Another straight line which passes through $$Q$$ and is perpendicular to the plane $$L_1$$ intersects the plane $$L_2 : 2x - y + z = -4$$ at the point $$R$$. Then which of the following statements is(are) TRUE?

The given point is $$P(1,3,2)$$ and the line parallel to which we must draw a line through $$P$$ is
$$\frac{x-2}{1}=\frac{y-4}{2}=\frac{z-6}{1}$$

The direction ratios of this line are $$1,2,1$$, so a line through $$P$$ with the same direction has parametric form
$$x = 1 + t,\; y = 3 + 2t,\; z = 2 + t \qquad -(1)$$

This line meets the plane $$L_1 : x - y + 3z = 6$$ at the point $$Q$$. Substitute the coordinates from $$(1)$$ into the plane:

$$\bigl(1+t\bigr) - \bigl(3+2t\bigr) + 3\bigl(2+t\bigr) = 6$$
$$1+t - 3 - 2t + 6 + 3t = 6$$
$$4 + 2t = 6 \;\;\Longrightarrow\;\; t = 1$$

Putting $$t=1$$ back in $$(1)$$ gives
$$Q(2,\,5,\,3)$$

Vector $$\overrightarrow{PQ} = (2-1,\,5-3,\,3-2) = (1,2,1)$$, so
$$PQ = \sqrt{1^{2}+2^{2}+1^{2}} = \sqrt{6}$$
Thus statement A is true.

A line through $$Q$$ perpendicular to plane $$L_1$$ must have direction equal to the normal of $$L_1$$.
Since $$L_1$$ is $$x - y + 3z = 6$$, its normal vector is $$(1,-1,3)$$.

Parametric equations of this perpendicular line are
$$x = 2 + s,\; y = 5 - s,\; z = 3 + 3s \qquad -(2)$$

This line meets the plane $$L_2 : 2x - y + z = -4$$ at point $$R$$. Insert the coordinates from $$(2)$$ into $$L_2$$:

$$2(2+s) - (5 - s) + (3 + 3s) = -4$$
$$4 + 2s - 5 + s + 3 + 3s = -4$$
$$2 + 6s = -4$$
$$6s = -6 \;\;\Longrightarrow\;\; s = -1$$

Substituting $$s=-1$$ in $$(2)$$ gives
$$R(1,\,6,\,0)$$

Thus statement B, which claimed $$R(1,6,3)$$, is false.

Centroid $$G$$ of $$\triangle PQR$$ is
$$G\bigl(\,\tfrac{1+2+1}{3},\;\tfrac{3+5+6}{3},\;\tfrac{2+3+0}{3}\bigr) = \Bigl(\tfrac{4}{3},\;\tfrac{14}{3},\;\tfrac{5}{3}\Bigr)$$

Hence statement C is true.

For statement D, compute the remaining two sides:

$$QR = \sqrt{(1)^{2} + (1)^{2} + (-3)^{2}} = \sqrt{11}$$
$$RP = \sqrt{(0)^{2} + (-3)^{2} + 2^{2}} = \sqrt{13}$$

Perimeter $$= PQ + QR + RP = \sqrt{6} + \sqrt{11} + \sqrt{13}$$, not $$\sqrt{2} + \sqrt{6} + \sqrt{11}$$, so statement D is false.

Therefore the correct statements are:
Option A and Option C.