Equation of st. line joining $Q(2,3,5)$ and $R(1,-1,4)$ is

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

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

Since $P$ also lies on $5 x-4 y-z=1$

$\therefore-5 \lambda+10+16 \lambda-12+\lambda-5=1$

$\Rightarrow 12 \lambda=8 \Rightarrow \lambda=\frac{2}{3} \quad \therefore P=\left(\frac{4}{3}, \frac{1}{3}, \frac{13}{3}\right)$

Now let another point $S$ on $Q R$ be

$(-\mu+2,-4 \mu+3,-\mu+5)$

Since $S$ is the foot of perpendicular drawn from

$T(2,1,4)$ to $Q R$, where dr's of $S T$ are $\mu, 4 \mu-2, \mu-1$ and dr's of $Q R$ are $-1,-4,-1$

$\begin{array}{ll}

\therefore & -\mu-16 \mu+8-\mu+1=0 \Rightarrow 18 \mu=9 \Rightarrow \mu=\frac{1}{2} \\

\therefore & S=\left(\frac{3}{2}, 1, \frac{9}{2}\right)

\end{array}$

$\therefore \quad$ Distance between $P$ and $S$

$\begin{array}{l}

=\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}}=\frac{1}{\sqrt{2}}

\end{array}$