Animated Solution for Mathematics - Three Dimensional Geometry: If the mirror image of the point $P(3,4,9)$ in the line $\frac{x-1}{3}=\frac{y+1}{2}=\frac{z-2}{1}$ is $(\alpha ,\beta ,\gamma )$, then $14(\alpha +\beta +\gamma )$ is :

We are tasked with finding the mirror image of a point $P(3,4,9)$ across a line $L$ defined by the equation:

Imagine a 3D space where point $P$ is reflected across line $L$ to reach point $Q(\alpha ,\beta ,\gamma )$. The line $L$ acts as the axis of symmetry for this reflection.

To find the reflection, we first determine the "foot of the perpendicular," denoted as point $N$, which lies on the line $L$ and is closest to $P$. We parameterize the line by setting the equation equal to $\lambda$:

The direction vector of the line is $\vec{b}=(3,2,1)$. For $N$ to be the foot of the perpendicular, the vector $\overrightarrow{PN}$ must be orthogonal to $\vec{b}$.

Calculating $\overrightarrow{PN}=N-P$, we obtain:

The condition for perpendicularity is the dot product $\overrightarrow{PN}\cdot \vec{b}=0$:

Expanding this expression yields:

Thus, we find the parameter value:

Substituting $\lambda =\frac{23}{14}$ back into the expression for $N$, we find the coordinates of the foot of the perpendicular:

Because $N$ is the midpoint between $P$ and its image $Q$, we use the midpoint formula $\frac{P+Q}{2}=N$:

Solving these equations for the coordinates of $Q$:

The problem requires the value of $14(\alpha +\beta +\gamma )$. Summing the coordinates:

Multiplying by 14:

The final result is 108.