If the image of the point $$P(1 , 2, a)$$ in the line $$ \frac{x-6}{3} = \frac{y - 7}{2} = \frac{7 -z}{2}$$ is $$Q(5, b, c)$$, then $$ a^{2}+b^{2}+c^{2}$$ is equal to

$$\frac{x-6}{3} = \frac{y-7}{2} = \frac{z-7}{-2}$$

Its direction vector is $$\vec{d} = (3, 2, -2)$$ and a point on the line is $$(6, 7, 7)$$.

For Q to be the image of P, the midpoint $$M = \frac{P+Q}{2}$$ must lie on the line, and $$\vec{PQ}$$ must be perpendicular to $$\vec{d}$$. The midpoint is $$M = \left(\frac{1+5}{2}, \frac{2+b}{2}, \frac{a+c}{2}\right) = \left(3, \frac{2+b}{2}, \frac{a+c}{2}\right)$$.

The condition that M lies on the line gives:
$$\frac{3-6}{3} = \frac{\frac{2+b}{2} - 7}{2} = \frac{\frac{a+c}{2} - 7}{-2}.$$
Since $$\frac{-3}{3} = -1$$, we have
$$\frac{\frac{2+b}{2} - 7}{2} = -1 \implies \frac{2+b}{2} - 7 = -2 \implies \frac{2+b}{2} = 5 \implies b = 8,$$
$$\frac{\frac{a+c}{2} - 7}{-2} = -1 \implies \frac{a+c}{2} - 7 = 2 \implies \frac{a+c}{2} = 9 \implies a + c = 18 \quad\text{...(i)}.$$

The vector $$\vec{PQ} = (5-1, 8-2, c-a) = (4, 6, c-a)$$ must be perpendicular to $$\vec{d} = (3, 2, -2)$$. Thus,
$$\vec{PQ} \cdot \vec{d} = 12 + 12 - 2(c-a) = 0 \implies 24 = 2(c-a) \implies c - a = 12 \quad\text{...(ii)}.$$

Solving (i) and (ii) together, $$a + c = 18$$ and $$c - a = 12$$, we add to get $$2c = 30 \implies c = 15$$, and hence $$a = 3$$

Finally, the required sum is
$$a^2 + b^2 + c^2 = 9 + 64 + 225 = 298.$$