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Q.
$ABC$ is a triangle in a plane with vertices $A(2, 3, 5), B(-1, 3, 2)$ and $C(\lambda , 5, \mu)$. If the median through $A$ is equally inclined to the
coordinate axes, then the value of $(\lambda^3 + \mu^3 + 5)$ is :
$D \equiv\left(\frac{-1+\lambda}{2}, 4, \frac{2+\mu}{2}\right)$
direction cosine of $AD =\left\{\frac{-1+\lambda}{2}-2,4-3, \frac{2+\mu}{2}-5\right\}$
$\left\{\frac{-1+\lambda}{2}-2,4-3, \frac{2+\mu}{2}-5\right\}$
$\overline{ AD }=\frac{\lambda-5}{2} i + j +\frac{\mu-8}{2} \hat{ k }$
$\Rightarrow \frac{\left(\frac{\lambda-5}{2}\right)}{\sqrt{\left(\frac{\lambda-5}{2}\right)^{2}+1^{2}+\left(\frac{\mu-8}{2}\right)^{2}}}=\frac{1}{\sqrt{\left(\frac{\lambda-5}{2}\right)^{2}+1+\left(\frac{\mu-8}{2}\right)^{2}}}=\frac{\left(\frac{\mu-8}{2}\right)}{\sqrt{\left(\frac{\lambda-5}{2}\right)^{2}+1+\left(\frac{\mu-8}{2}\right)^{2}}}$
$\overline{ AD } \cdot i =\overline{ AD } \cdot \hat{ j }=\overline{ AD } \cdot \hat{ k }$
$\lambda=7, \mu=10$
$\lambda^{3}+\mu^{3}+5=343+1000+5=1348$
