Distance from Point to Image in a Line

To find the sum of all possible values of $a$, we first determine the image of the point $P(1, 2, 7)$ in the given line: $\frac{x}{1} = \frac{y-1}{1} = \frac{z-2}{2} = \lambda$

Any general point $F$ on this line can be represented in terms of the parameter $\lambda$ as: $F(\lambda) = (\lambda, \lambda + 1, 2\lambda + 2)$

Let $F$ be the foot of the perpendicular from point $P(1, 2, 7)$ onto the line. Then the position vector of $F$ relative to $P$ is: $\vec{PF} = (\lambda - 1)\hat{i} + (\lambda + 1 - 2)\hat{j} + (2\lambda + 2 - 7)\hat{k} = (\lambda - 1)\hat{i} + (\lambda - 1)\hat{j} + (2\lambda - 5)\hat{k}$

Since $\vec{PF}$ is perpendicular to the direction vector of the line $\vec{d} = \hat{i} + \hat{j} + 2\hat{k}$, their scalar product must be zero: $\vec{PF} \cdot \vec{d} = 0$ $1(\lambda - 1) + 1(\lambda - 1) + 2(2\lambda - 5) = 0$ $\lambda - 1 + \lambda - 1 + 4\lambda - 10 = 0$ $6\lambda - 12 = 0 \implies \lambda = 2$

Substituting $\lambda = 2$ back into $F$, the coordinates of the foot of the perpendicular are: $F = (2, 2 + 1, 2(2) + 2) = (2, 3, 6)$

Let $I(x_0, y_0, z_0)$ be the image of point $P(1, 2, 7)$ in the line. The foot of the perpendicular $F$ is the midpoint of the line segment joining $P$ and $I$: $F = \frac{P + I}{2} \implies I = 2F - P$ $I = 2(2, 3, 6) - (1, 2, 7) = (4 - 1, 6 - 2, 12 - 7) = (3, 4, 5)$

Thus, the image of the point $(1, 2, 7)$ is $I(3, 4, 5)$.

We are given that the distance between the point $A(a, 2, 5)$ and the image $I(3, 4, 5)$ is equal to $4$: $\text{Distance}(A, I) = \sqrt{(a - 3)^2 + (2 - 4)^2 + (5 - 5)^2} = 4$

Squaring both sides: $(a - 3)^2 + (-2)^2 + 0^2 = 4^2$ $(a - 3)^2 + 4 = 16$ $(a - 3)^2 = 12$

Expanding the quadratic equation: $a^2 - 6a + 9 - 12 = 0$ $a^2 - 6a - 3 = 0$

By Vieta's formulas, the sum of all possible values of $a$ (the roots of the quadratic equation) is: $\text{Sum of values of } a = -\frac{-6}{1} = 6$

Therefore, the correct option is C.