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 ?

The direction ratios of line $L$ passing through $P(1, 2, -1)$ and $Q(2, 3, 1)$ are $(2-1, 3-2, 1-(-1)) = (1, 1, 2)$.

The equation of line $L$ is $\dfrac{x-1}{1} = \dfrac{y-2}{1} = \dfrac{z+1}{2} = \lambda$.

Any point on $L$ can be written as $S(\lambda+1, \lambda+2, 2\lambda-1)$.

Since $S$ is the foot of the perpendicular from $R(4, -1, 5)$ to $L$, the direction ratios of $RS$ are $(\lambda-3, \lambda+3, 2\lambda-6)$.

As $RS \perp L$, the dot product of their direction ratios is zero:

$1(\lambda-3) + 1(\lambda+3) + 2(2\lambda-6) = 0$

$6\lambda - 12 = 0 \Rightarrow \lambda = 2$

Thus, the coordinates of $S$ are $(3, 4, 3)$.

The length of $PS$ is $\sqrt{(3-1)^2 + (4-2)^2 + (3-(-1))^2} = \sqrt{4+4+16} = 2\sqrt{6}$.

The length of the altitude $RS$ is $\sqrt{(3-4)^2 + (4-(-1))^2 + (3-5)^2} = \sqrt{1+25+4} = \sqrt{30}$.

Since $S$ divides $PT$ internally in the ratio $1:2$, we have:

$S = \dfrac{T + 2P}{3} \Rightarrow T = 3S - 2P$

$T = 3(3, 4, 3) - 2(1, 2, -1) = (7, 8, 11)$

The length of the base $PT$ is $3|PS| = 6\sqrt{6}$.

The area of $\triangle PRT$ is $\dfrac{1}{2} \times |PT| \times |RS| = \dfrac{1}{2} \times 6\sqrt{6} \times \sqrt{30} = 18\sqrt{5}$.

The orthocentre $H$ lies on the altitude $RS$. The line $RS$ passes through $S(3, 4, 3)$ and has direction ratios proportional to $R-S = (1, -5, 2)$.

Let the coordinates of $H$ be $(3+t, 4-5t, 3+2t)$.

Since $H$ is the orthocentre, $PH \perp RT$.

The direction ratios of $RT$ are $(7-4, 8-(-1), 11-5) = (3, 9, 6)$, which is proportional to $(1, 3, 2)$.

The direction ratios of $PH$ are $(3+t-1, 4-5t-2, 3+2t-(-1)) = (t+2, 2-5t, 2t+4)$.

Taking the dot product of $PH$ and $RT$:

$1(t+2) + 3(2-5t) + 2(2t+4) = 0$

$t + 2 + 6 - 15t + 4t + 8 = 0$

$16 - 10t = 0 \Rightarrow t = \dfrac{8}{5}$

Substituting $t = \dfrac{8}{5}$ into the coordinates of $H$, we get:

$H = \left(3 + \dfrac{8}{5}, 4 - 5\left(\dfrac{8}{5}\right), 3 + 2\left(\dfrac{8}{5}\right)\right) = \left(\dfrac{23}{5}, -4, \dfrac{31}{5}\right)$

Answer: The orthocentre of the triangle $PRT$ is $\left(\dfrac{23}{5}, -4, \dfrac{31}{5}\right)$; The area of the triangle $PRT$ is $18\sqrt{5}$