Let $L$ be the straight line joining the points $P(1, 2, -1)$ and $Q(2, 3, 1)$. Let $S$ be the foot of the perpendicular drawn from the point $R(4, -1, 5)$ to the line $L$. Another line passing through $R$ intersects $L$ at a point $T$ such that the point $S$ divides the line segment $PT$ internally in the ratio $|PS| : |ST| = 1 : 2$, where $|PS|$ and $|ST|$ are the lengths of the line segments $PS$ and $ST$, respectively. Then which of the following statements is (are) TRUE ?

$L \text{ through } P(1,2,-1), \ Q(2,3,1)$ $\vec{PQ}=(1,1,2)$ $R(4,-1,5), \quad S = \text{foot of } \perp \text{ from } R \text{ to } L$ $T \in L, \quad R \text{ and } T \text{ joined by a line}$ $S \text{ divides } PT \text{ internally}: \ PS:ST=1:2$ $\text{Find: orthocentre and area of } \triangle PRT$ $P(1,2,-1), \quad Q(2,3,1)$ $\vec{d}=\overrightarrow{PQ}=Q-P=(1,1,2)$ $L:\ \vec{r}=P+t\vec{d}=(1,2,-1)+t(1,1,2)$ $R(4,-1,5), \quad \overrightarrow{PR}=R-P=(3,-3,6)$ $t=\dfrac{\overrightarrow{PR}\cdot\vec{d}}{|\vec{d}|^2}=\dfrac{3-3+12}{6}=2$ $S=P+2\vec{d}=(3,4,3)$ $PS:ST=1:2 \Rightarrow S=\dfrac{2P+T}{3}$ $T=3S-2P=(9,12,9)-(2,4,-2)=(7,8,11)$ $T=P+6\vec{d}\ \Rightarrow\ T\in L$ $|PS|=2\sqrt6,\quad |ST|=4\sqrt6 \Rightarrow PS:ST=1:2\ \checkmark$ $1.\ RS\perp PT \Rightarrow H\in \text{line } RS$ $2.\ \text{altitude from } P\perp RT \Rightarrow H$ $3.\ \text{Area}=\tfrac12|\overrightarrow{PR}\times\overrightarrow{PT}|$ $S(3,4,3), \quad \overrightarrow{SR}=R-S=(1,-5,2)$ $H=S+\lambda(1,-5,2)=(3+\lambda,\ 4-5\lambda,\ 3+2\lambda)$ $\overrightarrow{RT}=T-R=(3,9,6)$ $(\vec{H}-\vec{P})\cdot\overrightarrow{RT}=0$ $\vec{H}-\vec{P}=(2+\lambda,\ 2-5\lambda,\ 4+2\lambda)$ $(2+\lambda)3+(2-5\lambda)9+(4+2\lambda)6=0$ $48-30\lambda=0 \Rightarrow \lambda=\dfrac{8}{5}$ $H=\left(3+\dfrac{8}{5},\ 4-8,\ 3+\dfrac{16}{5}\right)$ $H=\left(\dfrac{23}{5},-4,\dfrac{31}{5}\right)$ $\text{Option (1) TRUE}, \quad \text{Option (2) FALSE}$ $\text{Area}=\dfrac12|\overrightarrow{PR}\times\overrightarrow{PT}|$ $\overrightarrow{PR}=(3,-3,6), \quad \overrightarrow{PT}=(6,6,12)$ $\overrightarrow{PR}\times\overrightarrow{PT}=\begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\3&-3&6\\6&6&12\end{vmatrix}$ $=\hat{i}(-72)-\hat{j}(0)+\hat{k}(36)=(-72,0,36)$ $|\overrightarrow{PR}\times\overrightarrow{PT}|=\sqrt{72^2+36^2}=\sqrt{6480}$ $=\sqrt{36^2\cdot 5}=36\sqrt5$ $\text{Area}=\dfrac12\cdot 36\sqrt5=18\sqrt5$ $\text{Option (4) TRUE}, \quad \text{Option (3) FALSE}$ $H=\left(\dfrac{23}{5},-4,\dfrac{31}{5}\right), \quad \text{Area}=18\sqrt5$ $\boxed{\text{Answer: (1) and (4)}}$