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

A 3D Geometry Odyssey

Welcome, fellow traveler of the JEE Advanced landscape. Today, we are not just solving a problem; we are witnessing a beautiful, symmetric dance in three-dimensional space.

Imagine you are standing in a room with two mirrors, $L_1$ and $L_2$. You have a point $P(a,2,a)$. You look into mirror $L_1$, and you see your reflection, $Q$.

But here is the twist: when you look at $Q$ in mirror $L_2$, you see yourself, $P$, staring back. This is a mutual reflection, a perfect cycle of symmetry. Let us unravel this mystery together.

The Anchor of Symmetry

When a point $P$ reflects to $Q$ across a line $L_1$, the line $L_1$ acts as the perpendicular bisector of the segment $PQ$. This means the midpoint $M$ of $PQ$ must lie on $L_1$.

Now, consider the second part of our journey: $Q$ reflects to $P$ across $L_2$. By the same logic, the midpoint of $QP$—which is the same point $M$—must lie on $L_2$.

This is our breakthrough! The midpoint $M$ is the intersection point of the two lines. It is the anchor of our entire coordinate system. We can parameterize $M$ using the equation of $L_1$: $\frac{x}{2}=\frac{y+a}{1}=\frac{z}{1}=\lambda$.

Thus, we define $M$ as:

The Perpendicularity Condition

Now, let us invoke the soul of reflection. The line segment $PQ$ is not just any line; it is the common normal to both $L_1$ and $L_2$. Because $PQ$ is perpendicular to $L_1$, the dot product of their direction vectors must be zero.

First, we find the coordinates of $Q$. Since $M$ is the midpoint, $M=\frac{P+Q}{2}$, which implies $Q=2M-P$. Substituting $P(a,2,a)$ and our $M$, we get:

The direction vector of $PQ$ is $\overrightarrow{PQ}=Q-P=(4\lambda -2a,2\lambda -2a-4,2\lambda -2a)$. The direction vector of $L_1$ is $\overrightarrow{d_1}=(2,1,1)$.

Setting $\overrightarrow{PQ}\cdot \overrightarrow{d_1}=0$, we get:

Expanding this, we find $12\lambda -8a-4=0$, which simplifies to our first golden equation:

The Algebraic Bridge

We have one equation, but two unknowns, $a$ and $\lambda$. We need more information. We know $M$ also lies on $L_2$. Let us set the equation of $L_2$ equal to a new parameter, $\mu$:

Substituting $M(2\lambda ,\lambda -a,\lambda )$ into this, we get three expressions for $\mu$:

1. $\frac{2\lambda -2b}{2}=\lambda -b=\mu$ 2. $\frac{(\lambda -a)-a}{1}=\lambda -2a=\mu$ 3. $\frac{\lambda +2b}{-5}=\mu$

Equating the first two, we see $\lambda -b=\lambda -2a$, which gives us the elegant result $b=2a$. Now, using the third expression, $\lambda +2b=-5\mu$.

Substituting $b=2a$ and $\mu =\lambda -2a$, we get $\lambda +4a=-5(\lambda -2a)$, which simplifies to $\lambda =a$.

We are almost there! Substitute $\lambda =a$ into our first equation, $3\lambda -2a-1=0$:

Since $b=2a$, we find $b=2$. The problem asks for $a+b$, and thus:

Look at that! Through the power of geometric visualization and systematic algebra, we have arrived at the answer: 3. You have successfully navigated the reflection, the perpendicularity, and the parameterization.