Answer

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**Hint:**This is a problem related to Three Dimensional Geometry. Before solving the 3D problems, you should know the general formulae like the equation of the plane, straight line in that plane, directional ratio, and Pythagoras theorem to find the distance between two points in the plane. All these formulae will be used in the solution.

**Complete step by step answer:**

As we know that the Cartesian equation of the straight line in a plane passing through two points $A({x_1},{y_1},{z_1}){\text{ and }}B({x_2},{y_2},{z_2})$ can be written as $\dfrac{{x - {x_1}}}{{{x_2} - {x_1}}} = \dfrac{{y - {y_1}}}{{{y_2} - {y_1}}} = \dfrac{{z - {z_1}}}{{{z_2} - {z_1}}}$, therefore the equation of the straight line $QR$ joining the two points $Q(2,3,5){\text{ and }}R(1, - 1,4)$ can be written as

$ \Rightarrow \dfrac{{x - 2}}{{1 - 2}} = \dfrac{{y - 3}}{{ - 1 - 3}} = \dfrac{{z - 5}}{{4 - 5}} = \lambda $

This can be simplified and written as

$ \Rightarrow \dfrac{{x - 2}}{{ - 1}} = \dfrac{{y - 3}}{{ - 4}} = \dfrac{{z - 5}}{{ - 1}} = \lambda $

Taking out the negative sign from the above expression, we will get the following expression,

$ \Rightarrow \dfrac{{x - 2}}{1} = \dfrac{{y - 3}}{4} = \dfrac{{z - 5}}{1} = \lambda {\text{ }}...................{\text{(1)}}$

So that any point on the line (1) can be written as,

$

\Rightarrow x = \lambda + 2, \\

\Rightarrow y = 4\lambda + 3, \\

\Rightarrow z = \lambda + 5 \\

$ ……………….. (2)

Since the line is with the plane $5x - 4y - z = 1$; hence it will satisfy the plane equation and in turn, will give the value of $\lambda $ in the following way,

$

\Rightarrow 5(\lambda + 2) - 4(4\lambda + 3) - (\lambda + 5) = 1 \\

\Rightarrow 5\lambda + 10 - 16\lambda - 12 - \lambda - 5 = 1 \\

\Rightarrow - 12\lambda - 7 = 1 \\

\Rightarrow \lambda = - \dfrac{8}{{12}} = - \dfrac{2}{3} \\

$

Therefore, putting the value of $\lambda $ in (2), we get,

$

\Rightarrow (x,y,z) \Leftrightarrow ( - \dfrac{2}{3} + 2, - \dfrac{8}{3} + 3, - \dfrac{2}{3} + 5) \\

\Rightarrow (x,y,z) \Leftrightarrow (\dfrac{4}{3},\dfrac{1}{3},\dfrac{{13}}{3}) \\

$

Thus, point $P$ is $(\dfrac{4}{3},\dfrac{1}{3},\dfrac{{13}}{3})$.

Again we can assume $S$ from equation (1) as

$

\Rightarrow \dfrac{{x - 2}}{1} = \dfrac{{y - 3}}{4} = \dfrac{{z - 5}}{1} = \alpha \\

{\text{Then,}} \\

\Rightarrow x = \alpha + 2,y = 4\alpha + 3,z = \alpha + 5 \\

$

And therefore, direction ratios for $TS$ with point $T(2,1,4)$ as

$

= (\alpha + 2 - 2,4\alpha + 3 - 1,\alpha + 5 - 4) \\

= (\alpha ,4\alpha + 2,\alpha + 1) \\

$

And direction ratios for $QR = (2 - 1,3 + 1,5 - 4)$

Which will be $QR = (1,4,1)$

Since, $TS$ and $QR$ are perpendicular to each other, therefore, their scalar product will be zero. Thus,

$

\Rightarrow \alpha .1 + (4\alpha + 2).4 + (\alpha + 1).1 = 0 \\

\Rightarrow \alpha + 16\alpha + 8 + \alpha + 1 = 0 \\

\Rightarrow 18\alpha + 9 = 0 \\

\Rightarrow \alpha = - \dfrac{9}{{18}} = - \dfrac{1}{2} \\

$

So, $

\Rightarrow S = ( - \dfrac{1}{2} + 2,4( - \dfrac{1}{2}) + 3,( - \dfrac{1}{2}) + 5) \\

\Rightarrow S = (\dfrac{3}{2},1,\dfrac{9}{2}) \\

$

Now we know the Pythagoras theorem to calculate the distance between two points in the plane can be represented in the following manner. Say, two points $A({x_1},{y_1},{z_1}){\text{ and }}B({x_2},{y_2},{z_2})$ are in the plane, and then the distance between them can be calculated as

$ \Rightarrow d = \sqrt {{{({x_2} - {x_1})}^2} + {{({y_2} - {y_1})}^2} + {{({z_2} - {z_1})}^2}} $

Therefore, length (or distance) of $PS$ is as below,

$ \Rightarrow P (\dfrac{4}{3},\dfrac{1}{3},\dfrac{{13}}{3})$ and $S(\dfrac{3}{2},1,\dfrac{9}{2})$,

\[

\Rightarrow PS = \sqrt {{{(\dfrac{3}{2} - \dfrac{4}{3})}^2} + {{(1 - \dfrac{1}{3})}^2} + {{(\dfrac{9}{2} - \dfrac{{13}}{3})}^2}} \\

\Rightarrow PS = \sqrt {{{(\dfrac{{9 - 8}}{6})}^2} + {{(\dfrac{{3 - 1}}{3})}^2} + {{(\dfrac{{27 - 26}}{6})}^2}} \\

\]

Now, simplifying the above expression, we get

\[

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

\Rightarrow PS = \sqrt {\dfrac{{1 + 16 + 1}}{{36}}} \\

\Rightarrow PS = \sqrt {\dfrac{{18}}{{36}}} = \dfrac{1}{{\sqrt 2 }} \\

\]

**Therefore, the correct answer is option (a).**

**Note:**

Solving 3D geometrical problems, formulae for scalar and vector products for points, length or distance calculation, straight-line equations, and other important standard equations must be known. These will help to attempt the problems.

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