The point \(P\) is the intersection of the straight line joining the points \(Q(2,3,5)\) and \(R(1,-1,4)\) with the plane \(5 x-4 y-\) \(z=1\). If \(S\) is the foot of the perpendicular drawn from the point \(T(2,1,4)\) to \(Q R\), then the length of the line segment \(P S\) is

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}\)