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 $\left|PS\right|:\left|ST\right|=1:2$, where $\left|PS\right|$ and $\left|ST\right|$ are the lengths of the line segments $PS$ and $ST$, respectively. Then which of the following statements is (are) TRUE?

To determine the correct statements, we will analyze the given line $L$ and the geometric properties of triangle $PRT$ step-by-step.

Step 1: Equation of line $L$
Line $L$ passes through the points $P(1,2,-1)$ and $Q(2,3,1)$.
The direction vector of line $L$ is given by:

$\overrightarrow{d}=Q-P=(2-1,3-2,1-(-1\left)\right)=(1,1,2)$

Any general point on line $L$ can be expressed in terms of a parameter $\lambda$ as:

$(1+\lambda ,2+\lambda ,-1+2\lambda )$

Step 2: Finding the foot of the perpendicular $S$
Since $S$ lies on line $L$, its coordinates are $S(1+\lambda ,2+\lambda ,-1+2\lambda )$.
The vector $\overrightarrow{RS}$ from $R(4,-1,5)$ to $S$ is:

$\overrightarrow{RS}=(1+\lambda -4,2+\lambda -(-1),-1+2\lambda -5)=(\lambda -3,\lambda +3,2\lambda -6)$

Since $RS$ is perpendicular to line $L$, the dot product $\overrightarrow{RS}·\overrightarrow{d}=0$:

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

$\lambda -3+\lambda +3+4\lambda -12=0\Rightarrow 6\lambda -12=0\Rightarrow \lambda =2$

Substituting $\lambda =2$, we get the coordinates of $S$:

$S=(1+2,2+2,-1+4)=(3,4,3)$

Step 3: Finding point $T$
We are given that $S$ divides $PT$ internally in the ratio $1:2$.
Let $T=(x_T,y_T,z_T)$ and $P=(1,2,-1)$. Using the section formula:

$S=\left(\frac{1·x_T+2·1}{1+2},\frac{1·y_T+2·2}{1+2},\frac{1·z_T+2·(-1)}{1+2}\right)=(3,4,3)$

Equating the coordinates:
$\frac{x_T+2}{3}=3\Rightarrow x_T=7$
$\frac{y_T+4}{3}=4\Rightarrow y_T=8$
$\frac{z_T-2}{3}=3\Rightarrow z_T=11$
So, $T=(7,8,11)$.

Step 4: Area of triangle $PRT$
In triangle $PRT$, $RS$ is perpendicular to the base $PT$. Therefore, height $h=\left|RS\right|$ and base length is $\left|PT\right|$.
Let's find the length $\left|PS\right|$:

$\left|PS\right|=\sqrt{(3-1{)}^2+(4-2{)}^2+(3-(-1){)}^2}=\sqrt{2^2+2^2+4^2}=\sqrt{4+4+16}=\sqrt{24}=2\sqrt{6}$

Since $\left|PS\right|:\left|ST\right|=1:2$, total length $\left|PT\right|=3\left|PS\right|=6\sqrt{6}$.
Next, calculate the height $\left|RS\right|$ using $R(4,-1,5)$ and $S(3,4,3)$:

$\left|RS\right|=\sqrt{(3-4{)}^2+(4-(-1){)}^2+(3-5{)}^2}=\sqrt{(-1{)}^2+5^2+(-2{)}^2}=\sqrt{1+25+4}=\sqrt{30}$

Thus, the area of triangle $PRT$ is:

$\text{Area}=\frac{1}{2}\times \text{base}\times \text{height}=\frac{1}{2}\times 6\sqrt{6}\times \sqrt{30}=3\sqrt{180}=3\times 6\sqrt{5}=18\sqrt{5}$

Step 5: Orthocentre of triangle $PRT$
The orthocentre $H$ is the intersection point of the altitudes of triangle $PRT$.
One altitude is line $RS$, so $H$ lies on line $RS$.
The equation of line $RS$ passing through $R(4,-1,5)$ with direction vector $\overrightarrow{RS}=(-1,5,-2)$ is:

$H=(4-k,-1+5k,5-2k)$

Also, altitude from vertex $P$ is perpendicular to vector $\overrightarrow{RT}$.
Vector $\overrightarrow{RT}=T-R=(7-4,8-(-1),11-5)=(3,9,6)$.
Vector $\overrightarrow{PH}=H-P=(4-k-1,-1+5k-2,5-2k-(-1\left)\right)=(3-k,5k-3,6-2k)$.
Since $\overrightarrow{PH}·\overrightarrow{RT}=0$:

$3(3-k)+9(5k-3)+6(6-2k)=0$
$9-3k+45k-27+36-12k=0$
$30k+18=0\Rightarrow k=-\frac{18}{30}=-\frac{3}{5}$

Substituting $k=-\frac{3}{5}$ into coordinates of $H$:

$x_H=4-\left(-\frac{3}{5}\right)=\frac{23}{5}$
$y_H=-1+5\left(-\frac{3}{5}\right)=-1-3=-4$
$z_H=5-2\left(-\frac{3}{5}\right)=5+\frac{6}{5}=\frac{31}{5}$

Hence, the orthocentre is $\left(\frac{23}{5},-4,\frac{31}{5}\right)$.

Therefore, the true statements are:
1. The orthocentre of the triangle PRT is $\left(\frac{23}{5},-4,\frac{31}{5}\right)$
2. The area of the triangle PRT is $18\sqrt{5}$