1. Key Concepts and Formulas

• Equation of a Line in 3D: A line passing through a point $\vec{a}$ with a direction vector $\vec{d}$ can be represented parametrically as $\vec{r} = \vec{a} + t\vec{d}$, where $t$ is a scalar parameter.
• Foot of the Perpendicular: If $X$ is the foot of the perpendicular from a point $P$ to a line $L$, then the vector $\vec{PX}$ is perpendicular to the direction vector of the line $L$. This implies their dot product is zero: $\vec{PX} \cdot \vec{d} = 0$.
• Image of a Point in a Line: If $X$ is the foot of the perpendicular from point $P$ to line $L$, and $Q$ is the image of $P$ in $L$, then $X$ is the midpoint of the line segment $PQ$. Therefore, $X = \frac{P+Q}{2}$, which can be rearranged to find the image $Q = 2X - P$.

2. Step-by-Step Solution

Step 1: Determine the Parametric Equation of the Line AB

• What we are doing: We need to define the line $AB$ in which the image of point $P$ is to be found. This involves identifying a point on the line and its direction vector.
• Why: A parametric equation allows us to represent any general point on the line using a single variable, which is essential for finding the foot of the perpendicular.
• Math: Given points $A(4,7,1)$ and $B(3,5,3)$. The direction vector of the line $AB$, denoted as $\vec{d}$, is found by subtracting the coordinates of $A$ from $B$: $\vec{d} = \vec{AB} = B - A = (3-4, 5-7, 3-1) = (-1, -2, 2)$ Using point $A(4,7,1)$ and the direction vector $\vec{d}(-1,-2,2)$, any general point $X$ on the line $AB$ can be represented parametrically as: $X(t) = A + t\vec{d} = (4,7,1) + t(-1,-2,2) = (4-t, 7-2t, 1+2t)$ Here, $t$ is a real parameter.

Step 2: Formulate the Vector PX

• What we are doing: We are creating a vector that connects the given point $P$ to a general point $X$ on the line $AB$.
• Why: This vector is crucial for applying the perpendicularity condition in the next step.
• Math: The given point is $P(1,0,3)$. A general point on the line $AB$ is $X(4-t, 7-2t, 1+2t)$. The vector $\vec{PX}$ is obtained by subtracting the coordinates of $P$ from $X$: $\vec{PX} = X - P = ((4-t)-1, (7-2t)-0, (1+2t)-3)$ $\vec{PX} = (3-t, 7-2t, 2t-2)$

Step 3: Determine the Value of the Parameter $t$ for the Foot of the Perpendicular

• What we are doing: We are finding the specific value of the parameter $t$ that corresponds to the foot of the perpendicular from $P$ to the line $AB$.
• Why: The line segment $PX$ is perpendicular to the line $AB$ if and only if their direction vectors are orthogonal. In 3D geometry, this means their dot product is zero.
• Math: The direction vector of line $AB$ is $\vec{d} = (-1, -2, 2)$. The vector $\vec{PX} = (3-t, 7-2t, 2t-2)$. Since $\vec{PX}$ is perpendicular to $\vec{d}$, their dot product must be zero: $\vec{PX} \cdot \vec{d} = 0$ $(3-t)(-1) + (7-2t)(-2) + (2t-2)(2) = 0$ $-3 + t - 14 + 4t + 4t - 4 = 0$ $9t - 21 = 0$ $9t = 21$ $t = \frac{21}{9} = \frac{7}{3}$

Step 4: Calculate the Coordinates of the Foot of the Perpendicular $X$

• What we are doing: We substitute the value of $t$ found in the previous step back into the parametric equation of point $X$.
• Why: This gives us the exact coordinates of the foot of the perpendicular, which is a necessary intermediate step to find the image.
• Math: Using $t = \frac{7}{3}$ in $X(t) = (4-t, 7-2t, 1+2t)$: $X_x = 4 - \frac{7}{3} = \frac{12-7}{3} = \frac{5}{3}$ $X_y = 7 - 2\left(\frac{7}{3}\right) = 7 - \frac{14}{3} = \frac{21-14}{3} = \frac{7}{3}$ $X_z = 1 + 2\left(\frac{7}{3}\right) = 1 + \frac{14}{3} = \frac{3+14}{3} = \frac{17}{3}$ So, the foot of the perpendicular is $X\left(\frac{5}{3}, \frac{7}{3}, \frac{17}{3}\right)$.

Step 5: Find the Coordinates of the Image Point $Q(\alpha, \beta, \gamma)$

• What we are doing: We use the midpoint property, which states that the foot of the perpendicular $X$ is the midpoint of the original point $P$ and its image $Q$.
• Why: This is the fundamental geometric property defining the image of a point in a line.
• Math: Let $Q(\alpha, \beta, \gamma)$ be the image of $P(1,0,3)$. The foot of the perpendicular $X\left(\frac{5}{3}, \frac{7}{3}, \frac{17}{3}\right)$ is the midpoint of $PQ$. Using the midpoint formula $X = \frac{P+Q}{2}$, we can express $Q$ as $Q = 2X - P$: For the x-coordinate: $\alpha = 2X_x - P_x = 2\left(\frac{5}{3}\right) - 1 = \frac{10}{3} - \frac{3}{3} = \frac{7}{3}$ For the y-coordinate: $\beta = 2X_y - P_y = 2\left(\frac{7}{3}\right) - 0 = \frac{14}{3}$ For the z-coordinate: $\gamma = 2X_z - P_z = 2\left(\frac{17}{3}\right) - 3 = \frac{34}{3} - \frac{9}{3} = \frac{25}{3}$ So, the image point is $Q\left(\frac{7}{3}, \frac{14}{3}, \frac{25}{3}\right)$.

Step 6: Calculate the Sum $\alpha+\beta+\gamma$

• What we are doing: We sum the individual coordinates of the image point $Q$.
• Why: This is the final value requested by the question.
• Math: $\alpha+\beta+\gamma = \frac{7}{3} + \frac{14}{3} + \frac{25}{3}$ $\alpha+\beta+\gamma = \frac{7+14+25}{3} = \frac{46}{3}$

3. Common Mistakes & Tips

• Incorrect Direction Vector: Always ensure the direction vector is correctly calculated (e.g., $B-A$ or $A-B$, but not just random points).
• Sign Errors in Dot Product: Pay close attention to negative signs when calculating the dot product, as a single error can propagate through the entire solution.
• Midpoint Formula Misapplication: Remember that the foot of the perpendicular is the midpoint of the original point and its image. A common mistake is to confuse $Q = 2X - P$ with $Q = X - P$ or similar incorrect relationships.
• Fraction Arithmetic: Be meticulous with arithmetic involving fractions to avoid calculation errors.

4. Summary To find the image of point $P(1,0,3)$ in the line joining $A(4,7,1)$ and $B(3,5,3)$, we first established the parametric equation of the line $AB$. We then found the foot of the perpendicular, $X$, from $P$ to the line by using the condition that the vector $\vec{PX}$ is orthogonal to the line's direction vector, leading to a dot product of zero. Solving for the parameter $t$ allowed us to determine the exact coordinates of $X$. Finally, knowing that $X$ is the midpoint of $P$ and its image $Q$, we used the midpoint formula to calculate the coordinates of $Q(\alpha, \beta, \gamma)$. Summing these coordinates yielded $\alpha+\beta+\gamma = \frac{46}{3}$.

5. Final Answer The final answer is $\boxed{\frac{46}{3}}$, which corresponds to option (A).