Distance Between Line Plane Intersection and Foot of Perpendicular

To find the length of the line segment $PS$, we first find the coordinates of points $P$ and $S$.

Step 1: Find the equation of the line $QR$

The line passes through the points $Q(2,3,5)$ and $R(1,-1,4)$. The direction ratios of the line $QR$ are: $\vec{d} = (2-1, 3-(-1), 5-4) = (1, 4, 1)$

Thus, any point on the line $QR$ can be represented parametrically in terms of a parameter $t$ as: $(x, y, z) = (2 + t, 3 + 4t, 5 + t)$

Step 2: Find the coordinates of point $P$

Point $P$ is the intersection of the line $QR$ with the plane: $5x - 4y - z = 1$

Substituting the parametric coordinates of the line into the plane equation: $5(2 + t) - 4(3 + 4t) - (5 + t) = 1$ $10 + 5t - 12 - 16t - 5 - t = 1$ $-7 - 12t = 1$ $-12t = 8 \implies t_P = -\frac{2}{3}$

Substituting $t_P = -\frac{2}{3}$ into the parametric form gives the coordinates of $P$: $P = \left(2 - \frac{2}{3}, 3 + 4\left(-\frac{2}{3}\right), 5 - \frac{2}{3}\right) = \left(\frac{4}{3}, \frac{1}{3}, \frac{13}{3}\right)$

Step 3: Find the coordinates of point $S$

Point $S$ is the foot of the perpendicular dropped from the point $T(2,1,4)$ onto the line $QR$. Let the coordinates of $S$ be $(2 + s, 3 + 4s, 5 + s)$ for some parameter $s$.

The vector $\vec{TS}$ is given by: $\vec{TS} = S - T = (2 + s - 2, 3 + 4s - 1, 5 + s - 4) = (s, 2 + 4s, 1 + s)$

Since $TS$ is perpendicular to the line $QR$, the dot product $\vec{TS} \cdot \vec{d} = 0$: $1(s) + 4(2 + 4s) + 1(1 + s) = 0$ $s + 8 + 16s + 1 + s = 0$ $18s + 9 = 0 \implies s_S = -\frac{1}{2}$

Substituting $s_S = -\frac{1}{2}$ into the parametric form gives the coordinates of $S$: $S = \left(2 - \frac{1}{2}, 3 + 4\left(-\frac{1}{2}\right), 5 - \frac{1}{2}\right) = \left(\frac{3}{2}, 1, \frac{9}{2}\right)$

Step 4: Calculate the distance $PS$

Since both $P$ and $S$ lie on the line $QR$ with direction vector $\vec{d} = (1, 4, 1)$, the distance between them can be directly calculated as: $PS = |t_P - s_S| \cdot \|\vec{d}\|$

We have: $|t_P - s_S| = \left| -\frac{2}{3} - \left(-\frac{1}{2}\right) \right| = \left| -\frac{2}{3} + \frac{1}{2} \right| = \frac{1}{6}$ $\|\vec{d}\| = \sqrt{1^2 + 4^2 + 1^2} = \sqrt{18} = 3\sqrt{2}$

Therefore: $PS = \frac{1}{6} \times 3\sqrt{2} = \frac{\sqrt{2}}{2} = \frac{1}{\sqrt{2}}$

Thus, the length of the line segment $PS$ is $\frac{1}{\sqrt{2}}$, which corresponds to option (A).