Animated Solution for Mathematics - Three Dimensional Geometry: The equation of the plane passing through the line of intersection of the planes $\vec{r}\cdot (\hat{i}+\hat{j}+\hat{k})=1$ and $\vec{r}\cdot (2\hat{i}+3\hat{j}-\hat{k})+4=0$ and parallel to the $x$-axis is:

Imagine you are standing in a vast, three-dimensional room. You have two flat, infinite sheets of paper, $P_1$ and $P_2$, intersecting at a sharp, clean angle. This intersection is not just a point; it is a line, a spine that holds these two planes together.

In our problem, we are looking for a third plane that passes through this exact spine. This is a classic scenario in JEE geometry, and the most powerful tool in your arsenal is the concept of the family of planes.

By writing the equation as $P_1+\lambda P_2=0$, we are essentially creating a mathematical hinge. As we vary the parameter $\lambda$, we are rotating our new plane around that fixed line of intersection. It is a beautiful, dynamic way to visualize the problem.

Parallelism as a Vector Condition

Now, we are given a specific constraint: our new plane must be parallel to the $x$-axis. This is where many students stumble, but let us look at it through the lens of the normal vector.

Every plane has a normal vector, a perpendicular arrow that defines its orientation. If our plane is parallel to the $x$-axis, then its normal vector must be perpendicular to that same axis.

Since the $x$-axis is defined by the unit vector $\hat{i}$, the condition for our plane to be parallel to it is simply $\vec{n}\cdot \hat{i}=0$. This means the $x$-component of our normal vector must be zero. It is an elegant, simple requirement that cuts through the complexity of the 3D space.

Let us execute this. We start with the Cartesian forms of our planes: $x+y+z-1=0$ $2x+3y-z+4=0$

Using our family of planes formula, we write:

By grouping the terms, we get:

The coefficient of $x$ is $(1+2\lambda )$. Setting this to zero, we find:

This is the magic value that locks our plane into the required orientation. Substituting $\lambda =-\frac{1}{2}$ back into our grouped equation, the $x$ term vanishes, and we are left with:

Multiplying by $-2$ gives us the clean, final equation: $y-3z+6=0$

Converting this back to vector form, we arrive at:

You have successfully navigated the intersection, applied the constraint, and solved for the plane. This is the essence of JEE mathematics: taking a complex 3D visualization and reducing it to a simple, elegant algebraic truth.