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?

$\text{Given: } P(1,3,2), \quad \vec{d} = (1,2,1)$ $L_1: x - y + 3z = 6$ $L_2: 2x - y + z = -4$ $\text{To find: which statements are TRUE?}$ $(\text{1})\ |PQ| = \sqrt{6} \qquad (\text{2})\ R = (1, 6, 3)$ $(\text{3})\ \text{centroid} \qquad (\text{4})\ \text{perimeter} = \sqrt{2}+\sqrt{6}+\sqrt{11}$ $\vec{r} = (1,3,2) + t(1,2,1)$ $\text{Plan: } 1.\ \text{line} \cap L_1 = Q, \quad 2.\ |PQ|$ $3.\ \text{line, through } Q, \perp L_1, \cap L_2 = R$ $4.\ \text{centroid and perimeter of } \triangle PQR$ $(1+t) - (3+2t) + 3(2+t) = 6$ $4 + 2t = 6 \Rightarrow t = 1$ $Q = (2,5,3)$ $\vec{PQ} = Q - P = (1,2,1)$ $|PQ| = \sqrt{1+4+1} = \sqrt{6} \Rightarrow (1) \text{ is TRUE}$ $\text{Line through } Q \perp L_1 \Rightarrow \vec{d}_2 = \vec{n}_1 = (1,-1,3)$ $\vec{r} = (2,5,3) + s(1,-1,3)$ $2(2+s) - (5-s) + (3+3s) = -4$ $2 + 6s = -4 \Rightarrow s = -1$ $R = (1,6,0)$ $R = (1,6,0) \neq (1,6,3)$ $\Rightarrow (2) \text{ is FALSE}$ $G = \left(\dfrac{1+2+1}{3},\ \dfrac{3+5+6}{3},\ \dfrac{2+3+0}{3}\right)$ $G = \left(\dfrac{4}{3},\ \dfrac{14}{3},\ \dfrac{5}{3}\right) \Rightarrow (3) \text{ is TRUE}$ $QR = |(1,-1,3)| = \sqrt{11}, \quad PR = |(0,-3,2)| = \sqrt{13}$ $\text{Perimeter} = \sqrt{6}+\sqrt{11}+\sqrt{13} \neq \sqrt{2}+\sqrt{6}+\sqrt{11}$ $\Rightarrow (4) \text{ is FALSE}$ $\text{Answer: } (1) \text{ and } (3)$ $|PQ| = \sqrt{6}, \quad G = \left(\dfrac{4}{3},\dfrac{14}{3},\dfrac{5}{3}\right)$