1. Interpretation of the problem

The image of point $P(1,0,3)$ in the line joining $A(4,7,1)$ and $B(3,5,3)$ means the reflection of $P$ in the line $AB$.

If $Q(\alpha,\beta,\gamma)$ is the reflection of $P$ in line $AB$, then the foot of the perpendicular from $P$ to line $AB$ is the midpoint of $PQ$.

So we will:

• find the foot of perpendicular $M$ from $P$ to line $AB$,
• then use $M$ as midpoint of $PQ$ to get $Q$.

2. Equation of the line through $A$ and $B$

Given: $A(4,7,1), \quad B(3,5,3)$

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

Hence a general point on line $AB$ is $M = A + t\vec d = (4-t,\,7-2t,\,1+2t)$

3. Foot of perpendicular from $P$ to line $AB$

Since $M$ is the foot of perpendicular, vector $\overrightarrow{PM}$ must be perpendicular to direction vector $\vec d$.

Now $P=(1,0,3)$ so $\overrightarrow{PM}=M-P=(4-t-1,\,7-2t-0,\,1+2t-3)=(3-t,\,7-2t,\,-2+2t)$

Perpendicularity condition: $\overrightarrow{PM}\cdot \vec d=0$

So, $(3-t)(-1)+(7-2t)(-2)+(-2+2t)(2)=0$

Compute: $-3+t-14+4t-4+4t=0$ $9t-21=0$ $t=\frac{21}{9}=\frac{7}{3}$

Thus, $M=\left(4-\frac73,\,7-2\cdot\frac73,\,1+2\cdot\frac73\right)$ $M=\left(\frac53,\,\frac73,\,\frac{17}3\right)$

4. Use midpoint relation to find reflected point $Q$

Since $M$ is midpoint of $P$ and $Q$, $M=\frac{P+Q}{2}$ which gives $Q=2M-P$

Now, $2M=\left(\frac{10}3,\,\frac{14}3,\,\frac{34}3\right)$

Therefore, $Q=\left(\frac{10}3-1,\,\frac{14}3-0,\,\frac{34}3-3\right)$ $Q=\left(\frac73,\,\frac{14}3,\,\frac{25}3\right)$

So, $\alpha=\frac73,\quad \beta=\frac{14}3,\quad \gamma=\frac{25}3$

5. Find $\alpha+\beta+\gamma$

$\alpha+\beta+\gamma=\frac73+\frac{14}3+\frac{25}3$ $=\frac{46}{3}$

6. Check with options

The correct option is: $\boxed{\frac{46}{3}}$ which is Option A.

7. Comparison with stored correct answer

Stored correct answer: A

Our derived answer: A

So they agree.