$$ABC$$ 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

We have the triangle $$ABC$$ with

$$A(2,\,3,\,5), \qquad B(-1,\,3,\,2), \qquad C(\lambda,\,5,\,\mu).$$

The median through $$A$$ meets the midpoint of the side $$BC$$. Let us first find this midpoint. Using the midpoint formula

Midpoint $$M\Bigl(x_M,\,y_M,\,z_M\Bigr)=\left(\dfrac{x_B+x_C}{2},\,\dfrac{y_B+y_C}{2},\,\dfrac{z_B+z_C}{2}\right),$$

we substitute the coordinates of $$B$$ and $$C$$:

$$x_M=\dfrac{-1+\lambda}{2},\qquad y_M=\dfrac{3+5}{2}=4,\qquad z_M=\dfrac{2+\mu}{2}.$$

Now we write the vector $$\overrightarrow{AM}$$ by subtracting the coordinates of $$A$$ from those of $$M$$:

$$\overrightarrow{AM}=\Bigl(x_M-2,\,y_M-3,\,z_M-5\Bigr).$$

Carrying out the subtraction term by term,

$$\begin{aligned} x\text{-component}&=\dfrac{-1+\lambda}{2}-2 =\dfrac{-1+\lambda-4}{2} =\dfrac{\lambda-5}{2},\\[4pt] y\text{-component}&=4-3=1,\\[4pt] z\text{-component}&=\dfrac{2+\mu}{2}-5 =\dfrac{2+\mu-10}{2} =\dfrac{\mu-8}{2}. \end{aligned}$$

Thus

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

The condition given is that this median is equally inclined to the three coordinate axes. For a vector $$\bigl(l,\,m,\,n\bigr)$$ the cosines of the angles it makes with the $$x-,y-,z-$$axes are, respectively,

$$\cos\alpha=\dfrac{l}{\sqrt{l^{2}+m^{2}+n^{2}}},\qquad \cos\beta=\dfrac{m}{\sqrt{l^{2}+m^{2}+n^{2}}},\qquad \cos\gamma=\dfrac{n}{\sqrt{l^{2}+m^{2}+n^{2}}}.$$

If the vector is equally inclined to all three axes, then $$\alpha=\beta=\gamma$$, and hence

$$\cos\alpha=\cos\beta=\cos\gamma.$$

This equality forces the numerators to be equal, so we must have

$$l=m=n \quad\text{or}\quad l=m=n=-k,$$

i.e. the three components are equal in magnitude and have the same sign. Applying this to $$\overrightarrow{AM}$$, we set

$$\dfrac{\lambda-5}{2}=1=\dfrac{\mu-8}{2}\quad$$ or $$\quad \dfrac{\lambda-5}{2}=-1=\dfrac{\mu-8}{2}.$$

We examine each possibility.

First possibility:

$$\dfrac{\lambda-5}{2}=1 \;\Longrightarrow\; \lambda-5=2 \;\Longrightarrow\; \lambda=7,$$

$$\dfrac{\mu-8}{2}=1 \;\Longrightarrow\; \mu-8=2 \;\Longrightarrow\; \mu=10.$$

Second possibility:

$$\dfrac{\lambda-5}{2}=-1 \;\Longrightarrow\; \lambda-5=-2 \;\Longrightarrow\; \lambda=3,$$

$$\dfrac{\mu-8}{2}=-1 \;\Longrightarrow\; \mu-8=-2 \;\Longrightarrow\; \mu=6.$$

Next, the problem asks for the value of $$\lambda^{3}+\mu^{3}+5$$. We compute this expression for both sets of values and see which one appears in the options.

For $$\lambda=7,\;\mu=10$$:

$$\begin{aligned} \lambda^{3}+\mu^{3}+5 &=7^{3}+10^{3}+5\\ &=343+1000+5\\ &=1348. \end{aligned}$$

For $$\lambda=3,\;\mu=6$$:

$$\begin{aligned} \lambda^{3}+\mu^{3}+5 &=3^{3}+6^{3}+5\\ &=27+216+5\\ &=248. \end{aligned}$$

The value $$248$$ does not appear among the given options, whereas $$1348$$ does. Therefore the admissible solution is $$\lambda=7,\;\mu=10$$, giving

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

Hence, the correct answer is Option 2.