Animated Solution for Mathematics - Three Dimensional Geometry: If the image of the point $P(1,2,a)$ in the line $\frac{x-6}{3}=\frac{y-7}{2}=\frac{7-z}{2}$ is $Q(5,b,c)$, then $a^2+b^2+c^2$ is equal to

A 3D Journey

In 3D coordinate geometry, finding the image of a point $P$ in a line $L$ is a precise geometric consequence of the laws of optics. The line $L$ acts as a mirror, and for $Q$ to be the true image of $P$, the line must be the perpendicular bisector of the segment $PQ$.

This implies two fundamental conditions: the midpoint $M$ of $PQ$ must lie on the line $L$, and the vector $\overrightarrow{PQ}$ must be perpendicular to the direction vector $\vec{d}$ of the line.

Before proceeding, we must be vigilant regarding the given line equation:

The $z$-term is written as $7-z$. In standard form, we require the variable to come first, such as $z-7$. We rewrite the expression as:

Now, the direction vector $\vec{d}$ is clearly $(3,2,-2)$. This adjustment is the difference between a correct solution and a common error.

We know that the midpoint $M$ of $PQ$ lies on the line. Given $P(1,2,a)$ and $Q(5,b,c)$, the midpoint $M$ is:

Since $M$ lies on the line, its coordinates must satisfy the line equation. Substituting $x=3$ into the first ratio:

This indicates that for point $M$, every ratio in the line equation must equal $-1$. This allows us to solve for $b$ and $c$:

We use the second geometric condition: $\overrightarrow{PQ}\perp \vec{d}$. The vector $\overrightarrow{PQ}$ is $(5-1,b-2,c-a)$, which simplifies to $(4,b-2,c-a)$.

Since the dot product of perpendicular vectors is zero, we compute:

Substituting $b=-4$:

Given $a+c=9$ and $c=a$, we find $2a=9$, so $a=4.5$ and $c=4.5$.

We have successfully navigated the geometry and the algebra. We found $a=4.5$, $b=-4$, and $c=4.5$.

The final step is to compute $a^2+b^2+c^2$:

By respecting the geometric constraints and carefully handling the algebra, we turned a complex 3D problem into a series of logical, manageable steps.