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

$\text{Given: } P = QR \cap (5x-4y-z=1)$

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

$S = \text{foot of } \perp \text{ from } T \text{ on } QR$

$\text{To find: } |PS|$

$(1)\ \frac{1}{\sqrt{2}} \qquad (2)\ \sqrt{2} \qquad (3)\ 2 \qquad (4)\ \sqrt{5}$

$1.\ \vec{d} = R - Q = \text{direction of } QR$ $2.\ QR \text{ in parametric form} \ \cap\ \text{plane} \Rightarrow P$ $3.\ S = \text{foot of } \perp \text{ from } T,\ \text{then } |PS|$ $\vec{d} = R - Q = (1-2,\ -1-3,\ 4-5)$ $\vec{d} = (-1,\ -4,\ -1)$ $\vec{r} = Q + \lambda\vec{d} = (2,3,5) + \lambda(-1,-4,-1)$ $= (2-\lambda,\ 3-4\lambda,\ 5-\lambda)$ $5(2-\lambda) - 4(3-4\lambda) - (5-\lambda) = 1$ $10-5\lambda - 12+16\lambda - 5+\lambda = 1$ $-7 + 12\lambda = 1 \Rightarrow \lambda = \dfrac{2}{3}$ $P = \left(2-\tfrac{2}{3},\ 3-\tfrac{8}{3},\ 5-\tfrac{2}{3}\right) = \left(\dfrac{4}{3},\ \dfrac{1}{3},\ \dfrac{13}{3}\right)$ $\vec{QT} = T - Q = (0,\ -2,\ -1)$ $\mu = \dfrac{\vec{QT}\cdot\vec{d}}{\vec{d}\cdot\vec{d}}$ $\vec{QT}\cdot\vec{d} = 0 + 8 + 1 = 9,\qquad \vec{d}\cdot\vec{d} = 1 + 16 + 1 = 18$ $\mu = \dfrac{9}{18} = \dfrac{1}{2}$ $S = Q + \tfrac{1}{2}\vec{d} = \left(\dfrac{3}{2},\ 1,\ \dfrac{9}{2}\right)$ $P,\ S \in QR \Rightarrow |PS| = |\lambda_P - \mu_S|\,|\vec{d}|$ $|\vec{d}| = \sqrt{(-1)^2 + (-4)^2 + (-1)^2} = \sqrt{18} = 3\sqrt{2}$ $|PS| = \left(\dfrac{2}{3} - \dfrac{1}{2}\right)\cdot 3\sqrt{2} = \dfrac{1}{6}\cdot 3\sqrt{2}$ $= \dfrac{\sqrt{2}}{2} = \dfrac{1}{\sqrt{2}}$ $\vec{PS} = P - S = \left(-\dfrac{1}{6},\ -\dfrac{2}{3},\ -\dfrac{1}{6}\right)$ $|\vec{PS}| = \sqrt{\dfrac{1}{36}+\dfrac{4}{9}+\dfrac{1}{36}} = \dfrac{1}{\sqrt{2}}$ $\boxed{\text{Answer: (1)}\quad \dfrac{1}{\sqrt{2}}}$