Animated Solution for Mathematics - Three Dimensional Geometry: Let the equation of the plane, that passes through the point $(1,4,-3)$ and contains the line of intersection of the planes $3x-2y+4z-7=0$ and $x+5y-2z+9=0$, be $\alpha x+\beta y+\gamma z+3=0$, then $\alpha +\beta +\gamma$ is equal to :

A Journey Through 3D Space

Imagine you are standing in a vast, empty room. In front of you, two massive, flat sheets of glass—our planes $P_1$ and $P_2$—are slicing through each other. Where they meet, they create a sharp, perfectly straight line known as the 'line of intersection.'

Now, imagine you need to place a third sheet of glass—our required plane—such that it is hinged exactly on that line of intersection, and it must also touch a specific point $A(1,4,-3)$ floating in the air. This is the essence of the problem we are solving today.

In the world of JEE Advanced, we don't just solve problems; we look for the most elegant path. When you see a plane passing through the intersection of two others, your mind should immediately jump to the 'Family of Planes' concept.

Think of it as a 'pencil of planes'—a collection of all possible planes that share the same hinge. The equation is beautifully simple: $P_1+\lambda P_2=0$. Here, $\lambda$ is the 'magic parameter.' By changing $\lambda$, you are essentially rotating your plane around the line of intersection until it hits the target point $A$.

Let's write down our machinery. We have:

Applying our formula, the equation of our required plane is:

This equation represents the entire family of planes passing through the intersection. But we need the one specific plane that contains point $A(1,4,-3)$. If the plane passes through $A$, then the coordinates of $A$ must satisfy the equation.

Precision is Key

First, let's substitute $x=1,y=4,z=-3$ into our equation and evaluate the first bracket:

Now, evaluate the second bracket:

Our equation now simplifies to a linear equation in $\lambda$:

Solving for $\lambda$ is straightforward:

Now that we have our $\lambda =\frac{2}{3}$, we plug it back into our family equation:

To make this look clean, let's multiply the entire equation by $3$ to clear the fraction:

Expanding these brackets gives us:

Combining like terms, we get:

Here is where many students stumble. The question asks us to match our equation to the form $\alpha x+\beta y+\gamma z+3=0$. Our constant term is $-3$, but the required form has $+3$.

We must multiply our entire equation by $-1$ to align them:

Now, by comparing coefficients, we find $\alpha =-11,\beta =-4,$ and $\gamma =-8$. The final step is to calculate the sum $\alpha +\beta +\gamma$:

And there you have it! The final answer is -23. We navigated the geometry, used the powerful family of planes, avoided the constant-term trap, and arrived at the solution.