Consider a pyramid OPQRS located in the first octant (x ≥ 0, y ≥ 0,...

O\left(0,0, 0\right) \to Origin$<br/>$P\left(3,0, 0\right) \to on x axis$<br/>$R\left(0,3, 0\right) \to on y axis$<br/>$Q\left(3,3, 0\right)$<br/>$T\left(\dfrac{3}{2}, \dfrac{3}{2}, 0\right), 5\left(\dfrac{3}{2}, \dfrac{3}{2}, 3\right)$<br ><img src="https://cdn-question-pool.Quantrex.app/pyq/jee_advanced/math_43_solut_paper1.jpg"/><br ><br/>Given OP = OR = 3 and OPQR is a square<br >$⇒ OQ = 3\sqrt{2} ⇒ OT = \dfrac{3}{\sqrt{2}}$and$ST = 3<br >Let θ be a angle between OQ & OS<br >Using\delta SOT, tan \theta = \dfrac{ST}{OT} = \sqrt{2} ⇒ \theta = \tan^{- 1} \sqrt{2}$<br/><img src="https://cdn-question-pool.Quantrex.app/pyq/jee_advanced/image_file_123.jpg"/><br/>Clearly, equation of plane containing triangle OQS is x - y = 0   as$O\left(0,0, 0\right), Q\left(3,3, 0\right), S\left(\dfrac{3}{2}, \dfrac{3}{2}, 3\right)lies on it$<br >$let ax + by + cz = d$<br >\left(0,0, 0\right) ⇒ d = 0$
\left(3,3, 0\right) ⇒ a + b = 0$<br >\left(\dfrac{3}{2}, \dfrac{3}{2}, 3\right) ⇒ \dfrac{3a}{2} + \dfrac{3b}{2} + 3c = 0$
$⇒ c = 0$
$⇒ b = - a$
$x - y = 0$
Also, length of perpendicular from P to the plane containing the triangle OQS is PT $= \dfrac{3}{\sqrt{2}}$ .
Also equation of RS is \overset{\to}{r} = 3\overset{^}{j} + t\left(\dfrac{3}{2}\overset{^}{i} - \dfrac{3}{2}\overset{^}{j} + 3\overset{^}{k}\right)
$= \left(\dfrac{3t}{2}, 3 - \dfrac{3t}{2}, 3t\right)$

Let M be foot of $⊥$ from origin on line passing through R,S
Let co - ordinates of $M = \left(\dfrac{3t}{2}, 3 - \dfrac{3t}{2}, 3t\right)$
$∵$ $\overset{\to}{OM} . \overset{\to}{RS} = 0$
$⇒$ $\dfrac{9}{4}t - \dfrac{3}{2}\left(3 - \dfrac{3t}{2}\right) + 9t = 0$
$⇒$ $\dfrac{9t}{2} + 9t = \dfrac{9}{2}$
$⇒$ $t = \dfrac{1}{3}$
$∴$ $M = \left(\dfrac{1}{2}, \dfrac{5}{2}, 1\right)$
$⇒$ $OM = \sqrt{\dfrac{1}{4} + \dfrac{25}{4} + 1} = \sqrt{\dfrac{30}{4}} = \sqrt{\dfrac{15}{2}}$