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?

$P(1,3,2),\quad \text{direction of } PQ=(1,2,1)$ $L_1:\ x-y+3z=6 \quad (Q = \text{line} \cap L_1)$ $L_2:\ 2x-y+z=-4$ $QR \perp L_1 \Rightarrow \text{direction } (1,-1,3);\quad R \in L_2$ $\text{Check: } (1)\ PQ=\sqrt{6},\qquad (2)\ R=(1,6,3)$ $(3)\ \text{centroid}=\left(\dfrac{4}{3},\dfrac{14}{3},\dfrac{5}{3}\right),\quad (4)\ \text{perimeter}=\sqrt2+\sqrt6+\sqrt{11}$ $\text{Plan: }1.\ \text{line } P+t(1,2,1)\ \cap L_1 = Q$ $2.\ \text{line } Q+s(1,-1,3)\ \cap L_2 = R$ $3.\ \text{find } PQ,\ R,\ \text{centroid},\ \text{perimeter}$ $P(1,3,2),\quad \text{direction } (1,2,1)\ \text{(given)}$ $L_1:\ x-y+3z=6,\qquad L_2:\ 2x-y+z=-4$ $\text{line: } (x,y,z)=(1,3,2)+t(1,2,1)$ $x=1+t,\quad y=3+2t,\quad z=2+t$ $\text{put in } L_1:\ (1+t)-(3+2t)+3(2+t)=6$ $4+2t=6 \Rightarrow t=1$ $Q(2,5,3)$ $PQ=|t|\cdot|(1,2,1)|=1\cdot\sqrt{6}=\sqrt{6}$ $(1)\ \textbf{TRUE}$ $QR \perp L_1 \Rightarrow \text{direction } (1,-1,3)$ $\text{line: } (x,y,z)=(2,5,3)+s(1,-1,3)$ $\text{put in } L_2:\ 2(2+s)-(5-s)+(3+3s)=-4$ $2+6s=-4 \Rightarrow s=-1$ $R(1,6,0)$ $R=(1,6,0)\ \neq\ (1,6,3)$ $(2)\ \textbf{FALSE}$ $\bar{x}=\dfrac{1+2+1}{3}=\dfrac{4}{3},\quad \bar{y}=\dfrac{3+5+6}{3}=\dfrac{14}{3},\quad \bar{z}=\dfrac{2+3+0}{3}=\dfrac{5}{3}$ $(3)\ \textbf{TRUE}$ $QR=|s|\cdot|(1,-1,3)|=1\cdot\sqrt{11}=\sqrt{11}$ $RP=|(0,3,-2)|=\sqrt{13}$ $\text{Perimeter}=\sqrt6+\sqrt{11}+\sqrt{13}\ \neq\ \sqrt2+\sqrt6+\sqrt{11}\ \Rightarrow\ (4)\ \textbf{FALSE}$ $\text{TRUE statements: } (1)\ \text{and}\ (3)$