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

$Q(2, 3, 5), \qquad R(1, -1, 4)$

$\Pi: \ 5x - 4y - z = 1, \qquad T(2, 1, 4)$

$P = QR \cap \Pi, \qquad S = \text{foot of } \perp \text{ from } T \text{ to } QR$

$\text{Find } PS$

$1.\ \text{direction of } QR$

$2.\ P = QR \cap \Pi$

$3.\ S = \text{foot from } T \text{ to } QR, \text{ then } PS$

$\vec{d} = R - Q = (1-2,\ -1-3,\ 4-5) = (-1, -4, -1)$

$QR: \ (x,y,z) = (2,3,5) + s(-1,-4,-1)$

$x = 2 - s, \quad y = 3 - 4s, \quad z = 5 - s$

$5(2-s) - 4(3-4s) - (5-s) = 1$

$-7 + 12s = 1 \Rightarrow s = \dfrac{2}{3}$

$P = \left(2-\dfrac{2}{3},\ 3-\dfrac{8}{3},\ 5-\dfrac{2}{3}\right) = \left(\dfrac{4}{3},\ \dfrac{1}{3},\ \dfrac{13}{3}\right)$

$S = Q + t\,\vec{d}, \qquad t = \dfrac{(T-Q)\cdot \vec{d}}{|\vec{d}|^2}$

$T - Q = (0, -2, -1), \qquad |\vec{d}|^2 = 1 + 16 + 1 = 18$

$(T-Q)\cdot \vec{d} = 0 + 8 + 1 = 9 \Rightarrow t = \dfrac{9}{18} = \dfrac{1}{2}$

$S = \left(2-\dfrac{1}{2},\ 3-2,\ 5-\dfrac{1}{2}\right) = \left(\dfrac{3}{2},\ 1,\ \dfrac{9}{2}\right)$

$P - S = \left(\dfrac{4}{3}-\dfrac{3}{2},\ \dfrac{1}{3}-1,\ \dfrac{13}{3}-\dfrac{9}{2}\right) = \left(-\dfrac{1}{6},\ -\dfrac{2}{3},\ -\dfrac{1}{6}\right)$

$PS = \sqrt{\dfrac{1}{36} + \dfrac{4}{9} + \dfrac{1}{36}} = \dfrac{1}{\sqrt2}$

$PS = \dfrac{1}{\sqrt2} \Rightarrow \text{option (1)}$