If the mirror image of the point $$(2, 4, 7)$$ in the plane $$3x - y + 4z = 2$$ is $$(a, b, c)$$, the $$2a + b + 2c$$ is equal to

We need to find the mirror image of point $$(2, 4, 7)$$ in the plane $$3x - y + 4z = 2$$.

The normal to the plane is $$\vec{n} = (3, -1, 4)$$ and the mirror image of point $$P(x_0, y_0, z_0)$$ in the plane $$ax + by + cz = d$$ is given by $$\frac{x - x_0}{a} = \frac{y - y_0}{b} = \frac{z - z_0}{c} = -\frac{2(ax_0 + by_0 + cz_0 - d)}{a^2 + b^2 + c^2}.$$

Here we compute $$ax_0 + by_0 + cz_0 - d = 3(2) - 1(4) + 4(7) - 2 = 6 - 4 + 28 - 2 = 28,$$ $$a^2 + b^2 + c^2 = 9 + 1 + 16 = 26,$$ so $$t = -\frac{2(28)}{26} = -\frac{56}{26} = -\frac{28}{13}.$$

Using this parameter, the coordinates of the image point are $$a = 2 + 3\left(-\frac{28}{13}\right) = 2 - \frac{84}{13} = \frac{26 - 84}{13} = -\frac{58}{13},$$ $$b = 4 + (-1)\left(-\frac{28}{13}\right) = 4 + \frac{28}{13} = \frac{52 + 28}{13} = \frac{80}{13},$$ $$c = 7 + 4\left(-\frac{28}{13}\right) = 7 - \frac{112}{13} = \frac{91 - 112}{13} = -\frac{21}{13}.$$

Finally, we compute $$2a + b + 2c = 2\left(-\frac{58}{13}\right) + \frac{80}{13} + 2\left(-\frac{21}{13}\right) = \frac{-116 + 80 - 42}{13} = \frac{-78}{13} = -6.$$

The answer is Option B: -6.