A B C is a triangle in a plane with vertices A 2 ,   3 ,   5 ,   B - 1 ,   3 ,   2…

Let coordinate of mid-point of $BC$ is point $D$, then 

$D\equiv \left(\frac{\lambda -1}{2}, 4, \frac{\mu +2}{2}\right)$

And, $A\equiv \left(2,3,5\right)$

Therefore, direction ratios of $AD$ are  

Hence, direction cosines of $AD$ are

Since, median $AD$ is making equal angles with coordinate axes, therefore

$\frac{\frac{\lambda -5}{2}}{\sqrt{{\left(\frac{\lambda -5}{2}\right)}^2+1+{\left(\frac{\mu -8}{2}\right)}^2}}=\frac{1}{\sqrt{{\left(\frac{\lambda -5}{2}\right)}^2+1+{\left(\frac{\mu -8}{2}\right)}^2}}=\frac{\frac{\mu -8}{2}}{\sqrt{{\left(\frac{\lambda -5}{2}\right)}^2+1+{\left(\frac{\mu -8}{2}\right)}^2}}$

$\Rightarrow \frac{\lambda -5}{2}=1=\frac{\mu -8}{2}$

$\Rightarrow$ $\lambda =7 \mathrm{a}\mathrm{n}\mathrm{d}\mathrm{ }\mathrm{ }\mathrm{\mu }=10$

$\therefore {\lambda }^3+{\mu }^3+5=1348$