If the equation of the plane passing through the line of intersection of the planes $$2x - y + z = 3$$, $$4x - 3y + 5z + 9 = 0$$ and parallel to the line $$\dfrac{x+1}{-2} = \dfrac{y+3}{4} = \dfrac{z-2}{5}$$ is $$ax + by + cz + 6 = 0$$, then $$a + b + c$$ is equal to

The plane through the intersection of $$2x-y+z-3=0$$ and $$4x-3y+5z+9=0$$ is:

$$ (2x-y+z-3) + \lambda(4x-3y+5z+9) = 0 $$

$$ (2+4\lambda)x + (-1-3\lambda)y + (1+5\lambda)z + (-3+9\lambda) = 0 $$

This plane is parallel to the line with direction ratios (-2, 4, 5), so the normal to the plane is perpendicular to this direction:

$$ -2(2+4\lambda) + 4(-1-3\lambda) + 5(1+5\lambda) = 0 $$

$$ -4-8\lambda-4-12\lambda+5+25\lambda = 0 $$

$$ -3+5\lambda = 0 \implies \lambda = \frac{3}{5} $$

Substituting $$\lambda = \frac{3}{5}$$:

$$(2+\frac{12}{5})x + (-1-\frac{9}{5})y + (1+3)z + (-3+\frac{27}{5}) = 0$$

$$\frac{22}{5}x - \frac{14}{5}y + 4z + \frac{12}{5} = 0$$

Multiplying by $$\frac{5}{2}$$:

$$ 11x - 7y + 10z + 6 = 0 $$

So $$a = 11$$, $$b = -7$$, $$c = 10$$.

$$ a + b + c = 11 - 7 + 10 = 14 $$

The correct answer is 14.