Let A(-3, 2) and B(-2, 1) be the vertices of a triangle ABC. If the centroid of this triangle lies on the line $$3x + 4y + 2 = 0$$, then the vertex C lies on the line :

We are given vertices A(-3, 2) and B(-2, 1) of triangle ABC. The centroid of the triangle lies on the line $$3x + 4y + 2 = 0$$. We need to find the line on which vertex C(x, y) lies.

The centroid G of a triangle is the average of the coordinates of its vertices. So, the coordinates of G are:

$$ G_x = \frac{x_A + x_B + x_C}{3} = \frac{-3 + (-2) + x}{3} = \frac{-5 + x}{3} $$

$$ G_y = \frac{y_A + y_B + y_C}{3} = \frac{2 + 1 + y}{3} = \frac{3 + y}{3} $$

Since the centroid lies on the line $$3x + 4y + 2 = 0$$, we substitute the coordinates of G into this equation:

$$ 3 \left( \frac{-5 + x}{3} \right) + 4 \left( \frac{3 + y}{3} \right) + 2 = 0 $$

Simplify each term:

$$ 3 \times \frac{-5 + x}{3} = -5 + x $$

$$ 4 \times \frac{3 + y}{3} = \frac{4(3 + y)}{3} $$

So the equation becomes:

$$ (-5 + x) + \frac{4(3 + y)}{3} + 2 = 0 $$

To eliminate the fraction, multiply every term by 3:

$$ 3 \times (-5 + x) + 3 \times \frac{4(3 + y)}{3} + 3 \times 2 = 0 \times 3 $$

Simplify:

$$ 3(-5 + x) + 4(3 + y) + 6 = 0 $$

Expand the expressions:

$$ 3 \times (-5) + 3 \times x + 4 \times 3 + 4 \times y + 6 = 0 $$

$$ -15 + 3x + 12 + 4y + 6 = 0 $$

Combine the constant terms:

$$ -15 + 12 = -3 $$

$$ -3 + 6 = 3 $$

So the equation is:

$$ 3x + 4y + 3 = 0 $$

This is the equation that vertex C(x, y) must satisfy. Comparing with the options:

A. $$4x + 3y + 5 = 0$$

B. $$3x + 4y + 3 = 0$$

C. $$4x + 3y + 3 = 0$$

D. $$3x + 4y + 5 = 0$$

Option B matches the derived equation. Hence, the correct answer is Option B.