Question:

1) Find $Q$: The line passing through $P(1, 3, 2)$ parallel to the given direction is: $$x = 1 + \lambda, \quad y = 3 + 2\lambda, \quad z = 2 + \lambda$$ Substitute into plane $L_1 : x - y + 3z = 6$: $$(1+\lambda) - (3+2\lambda) + 3(2+\lambda) = 6 \implies 2\lambda + 4 = 6 \implies \lambda = 1$$ So, $Q = (2, 5, 3)$. The length of $PQ$ is: $$PQ = \sqrt{(2-1)^2 + (5-3)^2 + (3-2)^2} = \sqrt{6} \quad \text{(Option A is True)}$$ 2) Find $R$: The line through $Q(2, 5, 3)$ perpendicular to $L_1$ is along the normal $\vec{n}_1 = (1, -1, 3)$: $$x = 2 + \mu, \quad y = 5 - \mu, \quad z = 3 + 3\mu$$ Substitute into plane $L_2 : 2x - y + z = -4$: $$2(2+\mu) - (5-\mu) + (3+3\mu) = -4 \implies 6\mu + 2 = -4 \implies \mu = -1$$ So, $R = (1, 6, 0)$ (Option B is False). 3) Centroid of $\triangle PQR$: $$G = \left( \dfrac{1+2+1}{3}, \dfrac{3+5+6}{3}, \dfrac{2+3+0}{3} \right) = \left( \dfrac{4}{3}, \dfrac{14}{3}, \dfrac{5}{3} \right) \quad \text{(Option C is True)}$$ 4) Perimeter of $\triangle PQR$: - $PQ = \sqrt{6}$ - $QR = \sqrt{(1-2)^2 + (6-5)^2 + (0-3)^2} = \sqrt{1+1+9} = \sqrt{11}$ - $PR = \sqrt{(1-1)^2 + (6-3)^2 + (0-2)^2} = \sqrt{0+9+4} = \sqrt{13}$ Perimeter is $\sqrt{6} + \sqrt{11} + \sqrt{13}$ (Option D is False). Thus, the correct options are A and C.