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?

$\text{Given: } L \text{ joins } P(1,2,-1) \text{ and } Q(2,3,1)$ $S = \text{foot of the } \perp \text{ from } R(4,-1,5) \text{ to } L$ $S \text{ divides } PT \text{ internally in the ratio } |PS|:|ST| = 1:2$ $\text{To check: which of (1)-(4) are TRUE}$ $\text{(1)/(2) orthocentre of } \triangle PRT, \qquad \text{(3)/(4) area of } \triangle PRT$ $1.\ \text{direction of } L,\ \text{foot } S$ $2.\ T \text{ from } PS:ST=1:2 \qquad 3.\ \text{area}=\tfrac12|\vec{PR}\times\vec{PT}|$ $4.\ \text{orthocentre} \Rightarrow \text{check (1)-(4)}$ $\vec d=\vec{PQ}=(1,1,2)$ $L:\ \vec r=(1,2,-1)+t(1,1,2)$ $S=(1+t,\ 2+t,\ -1+2t)$ $\vec{SR}\cdot\vec d=0,\qquad \vec{SR}=(3-t,\ -3-t,\ 6-2t)$ $(3-t)+(-3-t)+2(6-2t)=0$ $12-6t=0\ \Rightarrow\ t=2$ $S=(3,4,3)$ $|PS|=t\,|\vec d|=2\sqrt{6}\ \Rightarrow\ |ST|=2|PS|=4\sqrt{6}$ $\dfrac{PS}{PT}=\dfrac13 \Rightarrow S=P+\dfrac13(T-P)$ $T=3S-2P=(9,12,9)-(2,4,-2)=(7,8,11)$ $\vec{PR}=(3,-3,6),\qquad \vec{PT}=(6,6,12)$ $\text{Area}=\dfrac12\,|\vec{PR}\times\vec{PT}|$ $\vec{PR}\times\vec{PT}=\begin{vmatrix}\hat i&\hat j&\hat k\\3&-3&6\\6&6&12\end{vmatrix}=(-72,\ 0,\ 36)$ $|\vec{PR}\times\vec{PT}|=\sqrt{72^2+36^2}=36\sqrt{5}$ $\text{Area}=\dfrac12\cdot36\sqrt{5}=18\sqrt{5}\ \Rightarrow\ (4)\ \checkmark,\quad (3)\ \times$ $RS\perp L,\quad P,S,T\in L\ \Rightarrow\ RS\perp PT$ $\text{orthocentre } H \in RS$ $H=R+\lambda(S-R),\qquad S-R=(-1,5,-2)$ $H=(4-\lambda,\ -1+5\lambda,\ 5-2\lambda)$ $\vec{RT}=(3,9,6)\parallel(1,3,2)$ $\vec{PH}\cdot\vec{RT}=0$ $\vec{PH}=(3-\lambda,\ -3+5\lambda,\ 6-2\lambda)$ $6+10\lambda=0\ \Rightarrow\ \lambda=-\dfrac{3}{5}$ $H=\left(\dfrac{23}{5},\ -4,\ \dfrac{31}{5}\right)\ \Rightarrow\ (1)\ \checkmark,\quad (2)\ \times$ $\boxed{(1)\ \text{and}\ (4)\ \text{are TRUE}}$