If the image of the point $(4, 4, 3)$ in the line $\frac{x - 1}{2} = \frac{y - 2}{1} = \frac{z - 1}{3}$ is $(\alpha, \beta, \gamma)$, then $\alpha + \beta + \gamma$ is equal to

Step 1: Find a general point on the line

We start by setting the given equation of the line equal to a parameter $\lambda$. This allows us to express the coordinates of any point on the line in terms of $\lambda$.

Step 2: Determine the foot of the perpendicular

Let $A(4, 4, 3)$ be the given point and $P(1 + 2\lambda, 2 + \lambda, 1 + 3\lambda)$ be a general point on the line. The vector $\vec{AP}$ connects point A to point P. Since P is the foot of the perpendicular from A to the line, $\vec{AP}$ must be perpendicular to the direction vector of the line, $\vec{d} = (2, 1, 3)$. Their dot product must be zero.

Step 3: Solve for $\lambda$

We expand the dot product equation and solve for $\lambda$. This value of $\lambda$ will give us the coordinates of the foot of the perpendicular from point A to the line.

Step 4: Find the coordinates of the foot of the perpendicular

Substitute $\lambda = 1$ back into the general point $P(1 + 2\lambda, 2 + \lambda, 1 + 3\lambda)$ to find the coordinates of the foot of the perpendicular, P.

Step 5: Find the image point

Let the image of point $A(4, 4, 3)$ be $A'(\alpha, \beta, \gamma)$. The foot of the perpendicular P is the midpoint of the segment $AA'$. We use the midpoint formula to find the coordinates of $A'$.

Step 6: Calculate the sum of the coordinates of the image point

Finally, we sum the coordinates of the image point $(\alpha, \beta, \gamma)$ to get the required value.

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