Animated Solution for Mathematics - Three Dimensional Geometry: If the mirror image of the point $(2,4,7)$ in the plane $3x-y+4z=2$ is $(a,b,c)$, the $2a+b+2c$ is equal to :

A Journey into 3D Space

Imagine you are standing in a vast, dark room. In front of you, suspended in the void, is a point $P(2,4,7)$.

Below it, stretching infinitely in all directions, is a plane defined by the equation $3x-y+4z-2=0$. This plane acts like a perfect, infinite mirror.

Your task is to find the exact location of the reflection of point $P$ on the other side of this mirror. Let us call this image point $Q(a,b,c)$. This is not just a calculation; it is a dance of vectors and symmetry.

The Symmetry of the Normal

Before we touch a single number, let us visualize the physics of the situation. When you look into a mirror, the line of sight from your eye to your reflection is always perpendicular to the surface of the mirror.

In 3D geometry, this "line of sight" is the line passing through $P$ and $Q$. Because this line is perpendicular to the plane, its direction must be parallel to the normal vector of the plane.

The coefficients of $x,y,$ and $z$ in the plane equation give us this normal vector directly: $\vec{n}=3\hat{i}-\hat{j}+4\hat{k}$. This is the spine of our problem, and every step we take must respect this orientation.

The Magic Formula

In the heat of a JEE Advanced exam, we rely on the elegant, derived formula for the mirror image of a point $(x_1,y_1,z_1)$ in a plane $Ax+By+Cz+D=0$:

This formula is a masterpiece of efficiency. It encapsulates the entire geometric relationship into a single, solvable ratio.

The term on the right is the "magic constant"—it represents the scaled distance from the point to the plane, doubled to account for the reflection.

The Calculation

Let us breathe and substitute our values carefully. We have $P(2,4,7)$ and the plane $3x-y+4z-2=0$. Plugging these into our formula, we get:

Now, let us tackle the numerator. Inside the parentheses, we have $6-4+28-2$, which simplifies beautifully to $28$. Multiplying by the $-2$ outside, we get $-56$.

For the denominator, we calculate the magnitude squared of the normal vector: $3^2+(-1{)}^2+4^2=9+1+16=26$. Our ratio becomes $\frac{-56}{26}$, which simplifies to $-\frac{28}{13}$. This is our anchor.

Finding the Coordinates

Now, we equate each component to our anchor, $-\frac{28}{13}$.

For $a$:

For $b$:

For $c$:

We have found our point $Q\left(-\frac{58}{13},\frac{80}{13},-\frac{21}{13}\right)$.

The Final Evaluation

The problem asks for the value of $2a+b+2c$. Let us substitute our hard-earned coordinates:

And there it is. The complexity collapses into a clean, integer result: $-6$.

Remember, in 3D geometry, the math is just a language describing the shape of space. When you solve these, visualize the plane, see the normal vector, and trust the symmetry. You have mastered the reflection!