1. Coordinates of the midpoint of $BC$

The median through $A$ goes to the midpoint of $BC$.

Given: $A(2,3,5),\quad B(-1,3,2),\quad C(\lambda,5,\mu)$

Midpoint of $BC$ is $M\left(\frac{-1+\lambda}{2},\frac{3+5}{2},\frac{2+\mu}{2}\right)=\left(\frac{\lambda-1}{2},4,\frac{\mu+2}{2}\right)$

2. Direction ratios of the median $AM$

The vector along $AM$ is $\overrightarrow{AM}=\left(\frac{\lambda-1}{2}-2,\;4-3,\;\frac{\mu+2}{2}-5\right)$

So, $\overrightarrow{AM}=\left(\frac{\lambda-5}{2},\;1,\;\frac{\mu-8}{2}\right)$

Thus its direction ratios are proportional to $\lambda-5,\, 2,\, \mu-8$

3. Condition: equally inclined to the coordinate axes

If a line is equally inclined to the three coordinate axes, then its direction cosines have equal magnitudes: $|l|=|m|=|n|$

Hence its direction ratios are proportional to $\pm 1,\pm 1,\pm 1$

Since one direction ratio here is $2$, the others must have the same magnitude $2$. Therefore, $|\lambda-5|=2,\qquad |\mu-8|=2$

So, $\lambda=7 \text{ or } 3$ $\mu=10 \text{ or } 6$

4. Compute $\lambda^3+\mu^3+5$

Check the possible combinations:

• If $\lambda=7,\ \mu=10$: $\lambda^3+\mu^3+5=7^3+10^3+5=343+1000+5=1348$

• If $\lambda=7,\ \mu=6$: $343+216+5=564$

• If $\lambda=3,\ \mu=10$: $27+1000+5=1032$

• If $\lambda=3,\ \mu=6$: $27+216+5=248$

Among the given options, only $1348$ is present.

5. Conclusion

Therefore, the required value is $\boxed{1348}$ which corresponds to Option B.