If the three distinct lines $$x + 2ay + a = 0$$, $$x + 3by + b = 0$$ and $$x + 4ay + a = 0$$ are concurrent, then the point $$(a, b)$$ lies on a:

The three given lines are:

1. $$x + 2ay + a = 0$$

2. $$x + 3by + b = 0$$

3. $$x + 4ay + a = 0$$

Since the lines are concurrent, they all pass through a common point. For three lines to be concurrent, the determinant of their coefficients must be zero. The general form of a line is $$px + qy + r = 0$$. For the given lines:

- Line 1: coefficients are $$p_1 = 1$$, $$q_1 = 2a$$, $$r_1 = a$$

- Line 2: coefficients are $$p_2 = 1$$, $$q_2 = 3b$$, $$r_2 = b$$

- Line 3: coefficients are $$p_3 = 1$$, $$q_3 = 4a$$, $$r_3 = a$$

The determinant condition is:

$$ \begin{vmatrix} 1 & 2a & a \\ 1 & 3b & b \\ 1 & 4a & a \\ \end{vmatrix} = 0 $$

Expanding the determinant:

$$ = 1 \cdot \left( (3b)(a) - (b)(4a) \right) - 2a \cdot \left( (1)(a) - (b)(1) \right) + a \cdot \left( (1)(4a) - (3b)(1) \right) $$

Compute each part:

First part: $$ (3b)(a) - (b)(4a) = 3ab - 4ab = -ab $$

Second part: $$ -2a \cdot (a - b) = -2a(a - b) = -2a^2 + 2ab $$

Third part: $$ a \cdot (4a - 3b) = 4a^2 - 3ab $$

Summing them up:

$$ -ab + (-2a^2 + 2ab) + (4a^2 - 3ab) = -ab - 2a^2 + 2ab + 4a^2 - 3ab $$

Combine like terms:

For $$a^2$$ terms: $$ -2a^2 + 4a^2 = 2a^2 $$

For $$ab$$ terms: $$ -ab + 2ab - 3ab = (-1 + 2 - 3)ab = -2ab $$

So, the expression becomes:

$$ 2a^2 - 2ab = 0 $$

Factor out 2:

$$ 2(a^2 - ab) = 0 $$

Divide both sides by 2:

$$ a^2 - ab = 0 $$

Factor:

$$ a(a - b) = 0 $$

So, either $$a = 0$$ or $$a - b = 0$$ (i.e., $$a = b$$).

Now, check for distinct lines. If $$a = 0$$:

- Line 1: $$x + 2(0)y + 0 = 0$$ → $$x = 0$$

- Line 2: $$x + 3by + b = 0$$

- Line 3: $$x + 4(0)y + 0 = 0$$ → $$x = 0$$

Lines 1 and 3 are identical (both are $$x = 0$$), violating distinctness. Thus, $$a = 0$$ is invalid.

Therefore, $$a = b$$. The point $$(a, b)$$ satisfies $$b = a$$, meaning it lies on the straight line $$y = x$$.

Hence, the correct answer is Option C.