Animated Solution for Mathematics - Three Dimensional Geometry: Find the equation of the plane containing the line $2x-y+z-3=0,3x+y+z=5$ and at a distance of $\frac{1}{\sqrt{6}}$ from the point $(2,1,-1)$.

Imagine you are standing in a 3D coordinate system, holding two flat sheets of paper. These sheets are our planes, $P_1:2x-y+z-3=0$ and $P_2:3x+y+z-5=0$.

When these two planes intersect, they meet along a straight line. Think of this line as the spine of a book, and the planes as the pages.

Any plane that passes through this line is just another page in that same book. We are looking for a specific plane that sits at a precise distance of $\frac{1}{\sqrt{6}}$ from the point $(2,1,-1)$.

To capture every possible plane passing through that intersection line, we use the elegant 'family of planes' equation: $P_1+\lambda P_2=0$.

By substituting our given planes, we get:

This equation is our master key. The parameter $\lambda$ is the variable that allows us to rotate through all possible planes in this family. Our goal is to find the specific value of $\lambda$ that satisfies our distance constraint.

Before we can use the distance formula, we need to tidy up our equation. We expand the brackets and group the coefficients of $x$, $y$, and $z$:

Now, it looks like a standard plane equation $ax+by+cz+d=0$. This is the form we need to apply the distance formula.

We are given that the perpendicular distance from the point $(2,1,-1)$ to our target plane is $\frac{1}{\sqrt{6}}$. The distance formula is:

Substituting our point $(2,1,-1)$ and our coefficients into this formula gives us a seemingly daunting expression. However, the beauty of JEE problems lies in how they simplify.

When we plug the values into the numerator, we get $|(3\lambda +2)(2)+(\lambda -1)(1)+(\lambda +1)(-1)-(5\lambda +3)|$.

As we simplify this, the $\lambda$ terms combine: $(6+1-1-5)\lambda$ becomes $1\lambda$, and the constants $(4-1-1-3)$ become $-1$. The entire numerator collapses to $|\lambda -1|$.

Similarly, the denominator simplifies to $\sqrt{11{\lambda }^2+12\lambda +6}$. Now, our equation is:

To solve for $\lambda$, we square both sides to remove the modulus and the square root:

Cross-multiplying gives $6({\lambda }^2-2\lambda +1)=11{\lambda }^2+12\lambda +6$. Expanding and rearranging leads to $5{\lambda }^2+24\lambda =0$.

Factoring this, we find $\lambda =0$ or $\lambda =-\frac{24}{5}$. The value $\lambda =0$ returns our original plane $P_1$, while $\lambda =-\frac{24}{5}$ gives us our target plane.

Substituting $\lambda =-\frac{24}{5}$ back into our family equation and simplifying, we arrive at the final result: