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 5x - 4y - z = 1. If S is the foot of the perpendicular drawn from the point T(2, 1, 4) to QR, then the length of the line segment PS is

$1/\sqrt{2}$

The equation of line QR is

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

The coordinates of any point on QR are $(-r + 2, -4r + 3, -r + 5).$

If it lies on the plane $5x - 4y - z = 1,$ then

$5(-r + 2)-4(-4r + 3) - (-r + 5) = 1 $

$⇒ 12 r = 8 ⇒ r = 2/3 $

So, the coordinates of P are $\left(\frac{4}{3}, \frac{1}{3}, \frac{13}{3}\right)$.

$∴ TP = \sqrt{\left(2-\frac{4}{3}\right)^2+\left(1-\frac{1}{3}\right)^2+\left(4-\frac{13}{3}\right)^2}=\sqrt{\frac{4}{9}+\frac{4}{9}+\frac{1}{9}}= 1$

The direction ratios of TP are proportional to 2, 2, -1.

Let θ be the acute angle between QR and TP. Then,

$cos \theta = \frac{|2×(-1)+2×(-4)+(-1)(-1)|}{\sqrt{1+16+1}\sqrt{4+4+1}}=\frac{1}{\sqrt{2}}⇒ \theta = \frac{\pi}{4}$

∴ PS = ST

But, PT = 1 and $ \theta = \frac{\pi}{4}$, Therefore, PS = ST = $\frac{1}{\sqrt{2}}$