Let A(2, 3, 5), B($$-1, 3, 2$$) and C($$\lambda, 5, \mu$$) be the vertices of a $$\triangle$$ABC. If the median through A is equally inclined to the coordinate axes, then:

We are given the vertices of triangle ABC: A(2, 3, 5), B(-1, 3, 2), and C(λ, 5, μ). The median through vertex A is equally inclined to the coordinate axes. We need to find the relation between λ and μ that satisfies this condition.

First, recall that the median through A is the line segment joining A to the midpoint of the opposite side, which is BC. Let M be the midpoint of BC. The coordinates of M are found by averaging the coordinates of B and C:

$$ M_x = \frac{-1 + \lambda}{2} $$

$$ M_y = \frac{3 + 5}{2} = \frac{8}{2} = 4 $$

$$ M_z = \frac{2 + \mu}{2} $$

So, M is at $$ \left( \frac{-1 + \lambda}{2}, 4, \frac{2 + \mu}{2} \right) $$.

The median is the line segment AM, where A is (2, 3, 5) and M is $$ \left( \frac{-1 + \lambda}{2}, 4, \frac{2 + \mu}{2} \right) $$. The direction vector of AM is given by the vector from A to M:

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

Simplify each component:

First component: $$ \frac{-1 + \lambda}{2} - 2 = \frac{-1 + \lambda}{2} - \frac{4}{2} = \frac{-1 + \lambda - 4}{2} = \frac{\lambda - 5}{2} $$

Second component: $$ 4 - 3 = 1 $$

Third component: $$ \frac{2 + \mu}{2} - 5 = \frac{2 + \mu}{2} - \frac{10}{2} = \frac{2 + \mu - 10}{2} = \frac{\mu - 8}{2} $$

So, the direction vector is $$ \overrightarrow{AM} = \left( \frac{\lambda - 5}{2}, 1, \frac{\mu - 8}{2} \right) $$.

Since direction ratios are proportional, we can multiply the entire vector by 2 to simplify, giving direction ratios as:

$$ a = \lambda - 5, \quad b = 2, \quad c = \mu - 8 $$

The median being equally inclined to the coordinate axes means that the angles it makes with each axis are equal. This implies that the absolute values of the direction cosines are equal. For direction ratios a, b, c, the direction cosines are proportional to a, b, c, and the condition |l| = |m| = |n| requires |a| = |b| = |c|, because the denominator (the magnitude) is the same for all.

Given b = 2, we have |b| = 2. Therefore, |a| = 2 and |c| = 2:

$$ |\lambda - 5| = 2 \quad \text{and} \quad |\mu - 8| = 2 $$

Solving these equations:

For |λ - 5| = 2:

$$ \lambda - 5 = 2 \quad \text{or} \quad \lambda - 5 = -2 $$

$$ \lambda = 7 \quad \text{or} \quad \lambda = 3 $$

For |μ - 8| = 2:

$$ \mu - 8 = 2 \quad \text{or} \quad \mu - 8 = -2 $$

$$ \mu = 10 \quad \text{or} \quad \mu = 6 $$

This gives four possible pairs: (λ, μ) = (7, 10), (7, 6), (3, 10), (3, 6). However, for the median to be equally inclined, we must ensure that the direction cosines are equal (l = m = n), not just their absolute values. This requires the direction ratios to be equal: a = b = c.

So, set a = b and a = c:

$$ \lambda - 5 = 2 \quad \text{and} \quad \lambda - 5 = \mu - 8 $$

From λ - 5 = 2, we get λ = 7.

Substituting into λ - 5 = μ - 8: 7 - 5 = μ - 8 → 2 = μ - 8 → μ = 10.

Thus, the only pair that satisfies l = m = n is (λ, μ) = (7, 10).

Now, we check which option is satisfied by λ = 7 and μ = 10:

Option A: 5λ - 8μ = 5(7) - 8(10) = 35 - 80 = -45 ≠ 0

Option B: 8λ - 5μ = 8(7) - 5(10) = 56 - 50 = 6 ≠ 0

Option C: 10λ - 7μ = 10(7) - 7(10) = 70 - 70 = 0

Option D: 7λ - 10μ = 7(7) - 10(10) = 49 - 100 = -51 ≠ 0

Option C is satisfied. To verify, with λ = 7 and μ = 10, point C is (7, 5, 10). The midpoint M of BC is:

$$ M_x = \frac{-1 + 7}{2} = \frac{6}{2} = 3 $$

$$ M_y = \frac{3 + 5}{2} = \frac{8}{2} = 4 $$

$$ M_z = \frac{2 + 10}{2} = \frac{12}{2} = 6 $$

So M(3, 4, 6). The direction vector of AM is (3 - 2, 4 - 3, 6 - 5) = (1, 1, 1), which has direction ratios 1, 1, 1. The direction cosines are equal: l = m = n = 1/√3, confirming the median is equally inclined to the axes.

Hence, the correct answer is Option C.