Three urns A, B and C contain 4 red, 6 black, 5 red, 5 black, and $\lambda$  red, 4 black balls respectively. One of the urns is selected at random and a ball is drawn. If the ball drawn is red and the probability that it is drawn from urn C is 0.4, then the square of the length of the side of the largest equilateral triangle, inscribed in the parabola $y^2=\lambda x$ with one vertex at the vertex of the parabola is__

Urn       Red       Black
A             4             6
B             5             5
C             $\lambda$            4
Let E be the event to set red ball from ‘C’
$P(C/E)=\frac{P\left(C\right)\cdot P(E/C)}{P\left(A\right)P(E/A)+P\left(B\right)P(E/B)+P\left(C\right)P(E/C)}$

Given   $0.4=\frac{\frac{1}{3}\times \frac{\lambda }{\lambda +4}}{\frac{1}{3}(\frac{4}{10}+\frac{5}{10}+\frac{\lambda }{\lambda +4})}$

$$\begin{gathered} \frac{4}{10}(\frac{9}{10}+\frac{\lambda }{\lambda +4})=\frac{\lambda }{\lambda +4} \\ 76\lambda +144=100\lambda \Rightarrow 24\lambda =144\Rightarrow \lambda =6 \end{gathered}$$

Given parabola  $y^2=\lambda x \Rightarrow y^2=6x$

$$\begin{gathered} \sin {30}^0=\frac{BD}{a}\Rightarrow BD=\frac{a}{2} \\ \cos {30}^0=\frac{AD}{a}\Rightarrow AD=\frac{\sqrt{3}}{2}a \end{gathered}$$

$B=(\frac{\sqrt{3}}{2}a,\frac{a}{2})$lies on parabola   $y^2=6x$

$$\begin{gathered} \frac{a^2}{4}=6\left(\frac{\sqrt{3}a}{2}\right)\Rightarrow a=12\sqrt{3} \\ {a}^2=144\times 3\Rightarrow a^2=432 \end{gathered}$$