If the equation of the plane passing through the line of intersection of the planes $$2x - 7y + 4z - 3 = 0$$, $$3x - 5y + 4z + 11 = 0$$ and the point $$(-2, 1, 3)$$ is $$ax + by + cz - 7 = 0$$, then the value of $$2a + b + c - 7$$ is ________.

The equation of the plane passing through the line of intersection of the planes $$2x - 7y + 4z - 3 = 0$$ and $$3x - 5y + 4z + 11 = 0$$ can be written as:

$$(2x - 7y + 4z - 3) + \lambda(3x - 5y + 4z + 11) = 0$$

Expanding, we get $$(2 + 3\lambda)x + (-7 - 5\lambda)y + (4 + 4\lambda)z + (-3 + 11\lambda) = 0$$.

Since this plane passes through the point $$(-2, 1, 3)$$, we substitute:

$$(2 + 3\lambda)(-2) + (-7 - 5\lambda)(1) + (4 + 4\lambda)(3) + (-3 + 11\lambda) = 0$$

Computing the constant terms: $$-4 - 7 + 12 - 3 = -2$$. Computing the coefficient of $$\lambda$$: $$-6 - 5 + 12 + 11 = 12$$.

So $$-2 + 12\lambda = 0$$, which gives $$\lambda = \dfrac{1}{6}$$.

Substituting $$\lambda = \dfrac{1}{6}$$ back:

$$x$$-coefficient: $$2 + \dfrac{3}{6} = \dfrac{5}{2}$$
$$y$$-coefficient: $$-7 - \dfrac{5}{6} = -\dfrac{47}{6}$$
$$z$$-coefficient: $$4 + \dfrac{4}{6} = \dfrac{14}{3}$$
Constant: $$-3 + \dfrac{11}{6} = -\dfrac{7}{6}$$

Multiplying the entire equation by 6: $$15x - 47y + 28z - 7 = 0$$.

We can verify: $$15(-2) - 47(1) + 28(3) - 7 = -30 - 47 + 84 - 7 = 0$$ ✓

Comparing with $$ax + by + cz - 7 = 0$$, we get $$a = 15$$, $$b = -47$$, $$c = 28$$.

Therefore, $$2a + b + c - 7 = 2(15) + (-47) + 28 - 7 = 30 - 47 + 28 - 7 = 4$$.

So, the answer is $$\boxed{4}$$.