Animated Solution for Mathematics - Three Dimensional Geometry: Let $Q(a,b,c)$ be the image of the point $P(3,2,1)$ in the line $\frac{x-1}{1}=\frac{y}{2}=\frac{z-1}{1}$. Then the distance of $Q$ from the line $\frac{x-9}{3}=\frac{y-9}{2}=\frac{z-5}{-2}$ is

Welcome, future engineers! Today, we are embarking on a journey through the elegant world of 3D coordinate geometry. This problem tests your ability to visualize spatial relationships and apply vector algebra with precision.

We are tracing the path of a point $P(3,2,1)$ as it reflects across a line $L_1$ to form an image $Q(a,b,c)$, and then measuring the distance of $Q$ from a second line $L_2$.

To find the reflection, we first identify the foot of the perpendicular $M$ from $P$ to the line $L_1$. The line $L_1$ is given by:

Any point $M$ on this line can be expressed in parametric form as $M(1+\lambda ,2\lambda ,1+\lambda )$. The direction vector of the line is $\overrightarrow{d_1}=(1,2,1)$.

The vector $\overrightarrow{PM}$ is defined as $M-P$:

Since $\overrightarrow{PM}$ is perpendicular to $\overrightarrow{d_1}$, their dot product must be zero:

Expanding this equation yields:

Substituting $\lambda =1$ back into the parametric form, we find the foot of the perpendicular $M$ is $(2,2,2)$.

Since $M$ is the midpoint of the segment $PQ$, we use the midpoint formula $M=\frac{P+Q}{2}$, which rearranges to $Q=2M-P$.

Substituting the coordinates of $M$ and $P$:

The reflected point is $Q(1,2,3)$.

We now calculate the distance from $Q(1,2,3)$ to the line $L_2:\frac{x-9}{3}=\frac{y-9}{2}=\frac{z-5}{-2}$. This line passes through $A(9,9,5)$ with direction vector $\overrightarrow{d_2}=(3,2,-2)$.

First, we define the vector $\overrightarrow{AQ}=Q-A=(-8,-7,-2)$. The distance $d$ is given by the formula $d=\frac{|\overrightarrow{AQ}\times \overrightarrow{d_2}|}{|\overrightarrow{d_2}|}$.

Calculating the cross product $\overrightarrow{AQ}\times \overrightarrow{d_2}$:

The magnitude of the cross product is:

The magnitude of the direction vector $\overrightarrow{d_2}$ is:

Finally, the distance $d$ is:

The final distance is $7$ units.