Equation of line QR is

$$\frac{x-2}{1}=\frac{y-3}{4}=\frac{z-5}{1}=\lambda$$

General point on QR will be $$(\lambda+2,4 \lambda+3, \lambda+5)$$

Let $$P \equiv(\lambda+2,4 \lambda+3, \lambda+5)$$

Since, point P also lies on the plane.

$$\begin{array}{rrrl} \therefore & 5(\lambda+2)-4(4 \lambda+3)-(\lambda+5) & =1 \\ \Rightarrow & 5 \lambda+10-16 \lambda-12-\lambda-5 & =1 \\ \Rightarrow & -12 \lambda-7 & =1 \end{array}$$

$$\begin{array}{rrrl} \Rightarrow & \lambda &=\frac{-8}{12} \\ \Rightarrow & \lambda &=\frac{-2}{3} \end{array}$$

Hence,

$$\begin{aligned} P & \equiv\left(\frac{-2}{3}+2,4\left(\frac{-2}{3}\right)+3, \frac{-2}{3}+5\right) \\ & \equiv\left(\frac{4}{3}, \frac{1}{3}, \frac{13}{3}\right) \end{aligned}$$

Now, for TS to be perpendicular to QR,

$$ \begin{array}{rlrl} & \lambda+4(4 \lambda+2)+(\lambda+1) & =0 \\ \Rightarrow & \lambda \lambda+16 \lambda+8+\lambda+1 & =0 \\ \Rightarrow & 18 \lambda =-9 \\ \Rightarrow & \lambda =\frac{-1}{2} \end{array}$$

$$\begin{aligned} \therefore \quad \text { Point } S & \equiv\left(\frac{-1}{2}+2,4\left(\frac{-1}{2}\right)+3, \frac{-1}{2}+5\right) \\ & \equiv\left(\frac{3}{2}, 1, \frac{9}{2}\right) \end{aligned}$$

$$\therefore$$ Length of line segment PS

$$\begin{aligned} & =\sqrt{\left(\frac{4}{3}-\frac{3}{2}\right)^2+\left(\frac{1}{3}-1\right)^2+\left(\frac{13}{3}-\frac{9}{2}\right)^2} \\ & =\sqrt{\frac{1}{36}+\frac{4}{9}+\frac{1}{36}} \\ & =\sqrt{\frac{1+16+1}{36}} \\ & =\sqrt{\frac{18}{36}} \\ & =\frac{1}{\sqrt{2}} \end{aligned}$$