The urns A, B and C contains 4 red, 6 black; 5 red, 5 black and λ...

Let $R$ be the event of drawing red balls.

$P\left(\dfrac{R}{A}\right) = \dfrac{4}{10}$

$P\left(\dfrac{R}{B}\right) = \dfrac{5}{10}$

$P\left(\dfrac{R}{C}\right) = \dfrac{\lambda}{4 + \lambda}$

Now,

$P\left(\dfrac{C}{R}\right) = 0 . 4$

$⇒ \dfrac{P\left(C\right)P\left(\dfrac{R}{C}\right)}{P\left(A\right)P\left(\dfrac{R}{A}\right) + P\left(B\right)P\left(\dfrac{R}{B}\right) + P\left(C\right)P\left(\dfrac{R}{C}\right)} = \dfrac{4}{10}$

$⇒ \dfrac{\dfrac{1}{3} \times \left(\dfrac{\lambda}{\lambda + 4}\right)}{\dfrac{1}{3} \times \dfrac{4}{10} + \dfrac{1}{3} \times \dfrac{5}{10} + \dfrac{1}{3} \times \left(\dfrac{\lambda}{\lambda + 4}\right)} = \dfrac{4}{10}$

$⇒ \dfrac{\left(\dfrac{\lambda}{\lambda + 4}\right)}{\dfrac{9}{10} + \left(\dfrac{\lambda}{\lambda + 4}\right)} = \dfrac{4}{10}$

$⇒ \dfrac{\lambda}{\lambda + 4} = \dfrac{4}{10}\left[\dfrac{9}{10} + \left(\dfrac{\lambda}{\lambda + 4}\right)\right]$

$⇒ \dfrac{5\lambda}{\lambda + 4} = 2\left[\dfrac{9}{10} + \left(\dfrac{\lambda}{\lambda + 4}\right)\right]$

$⇒ \dfrac{5\lambda}{\lambda + 4} = \dfrac{19\lambda + 36}{5\left(\lambda + 4\right)}$

$⇒ 25\lambda = 19\lambda + 36$

$⇒ 6\lambda = 36$

$⇒ \lambda = 6$

So, parabola is

${y}^{2} = 6x = 4 \times \dfrac{2}{3} \times x$

Comparing with ${y}^{2} = 4ax$, we get

$a = \dfrac{2}{3}$

Hence, we have

Let $P ≡ \left(\dfrac{2}{3}{t}^{2}, \dfrac{4}{3}t\right)$

In $∆ OPQ$, we have

$\tan \left(30 °\right) = \dfrac{2at}{a{t}^{2}} = \dfrac{2}{t}$

$⇒ \dfrac{2}{t} = \dfrac{1}{\sqrt{3}}$

$⇒ t = 2\sqrt{3}$

Side length of triangle$= 4at = 12\sqrt{3}$

Square of side of triangle

$= {\left(12\sqrt{3}\right)}^{2} = 432$ units