If the image of the point $$P(1, 0, 3)$$ in the line joining the points $$A(4, 7, 1)$$ and $$B(3, 5, 3)$$ is $$Q(\alpha, \beta, \gamma)$$, then $$\alpha + \beta + \gamma$$ is equal to

The line joining $$A(4,7,1)$$ and $$B(3,5,3)$$ will act as the mirror line.
For the image $$Q$$ of $$P(1,0,3)$$ in this line, the line $$AB$$ must be the perpendicular bisector of the segment $$PQ$$. Hence:

1. Find the foot of the perpendicular from $$P$$ to line $$AB$$ (call this foot $$M$$).
2. Use the midpoint condition $$M=\dfrac{P+Q}{2}$$ to get $$Q$$.
3. Add the coordinates of $$Q$$.

Step 1: Equation of line $$AB$$
Direction vector $$\vec d = B-A = (3-4,\,5-7,\,3-1)=(-1,-2,2)$$.

Parametric form: $$\vec r = (4,7,1)+t(-1,-2,2)$$, so any point $$M$$ on the line is $$M(4-t,\;7-2t,\;1+2t)$$.

Step 2: Foot of perpendicular $$M$$ from $$P$$
For $$PM$$ to be perpendicular to $$\vec d$$, we need$$(\vec{M}-\vec{P})\cdot\vec d=0.$$

Compute $$\vec{M}-\vec{P}=(4-t-1,\;7-2t-0,\;1+2t-3)=(3-t,\;7-2t,\;-2+2t).$$

Dot product with $$\vec d=(-1,-2,2):$$
$$(3-t)(-1)+(7-2t)(-2)+(-2+2t)(2)=0.$$

Simplify:
$$-(3-t)-2(7-2t)+2(-2+2t)=0$$
$$(-3+t)+(-14+4t)+(-4+4t)=0$$
$$-21+9t=0 \;\Longrightarrow\; t=\dfrac{21}{9}=\dfrac{7}{3}.$$

Thus $$M\Bigl(4-\dfrac{7}{3},\;7-\dfrac{14}{3},\;1+\dfrac{14}{3}\Bigr)=\Bigl(\dfrac{5}{3},\;\dfrac{7}{3},\;\dfrac{17}{3}\Bigr).$$

Step 3: Coordinates of the image $$Q(\alpha,\beta,\gamma)$$
Since $$M$$ is the midpoint of $$P$$ and $$Q$$,
$$Q = 2M - P.$$

Compute each coordinate:
$$\alpha = 2\!\left(\dfrac{5}{3}\right) - 1 = \dfrac{10}{3}-\dfrac{3}{3}=\dfrac{7}{3},$$ $$\beta = 2\!\left(\dfrac{7}{3}\right) - 0 = \dfrac{14}{3},$$ $$\gamma = 2\!\left(\dfrac{17}{3}\right) - 3 = \dfrac{34}{3}-\dfrac{9}{3}=\dfrac{25}{3}.$$

Step 4: Required sum
$$\alpha+\beta+\gamma = \dfrac{7}{3}+\dfrac{14}{3}+\dfrac{25}{3}=\dfrac{46}{3}.$$

Therefore, $$\alpha + \beta + \gamma = \dfrac{46}{3}$$, which matches Option B.