Animated Solution for Mathematics - Three Dimensional Geometry: A plane $P$ contains the line $x+2y+3z+1=0=x-y-z-6$, and is perpendicular to the plane $-2x+y+z+8=0$. Then which of the following points lies on $P$?

We begin with two planes, $P_1:x+2y+3z+1=0$ and $P_2:x-y-z-6=0$. They meet at a line.

Any plane passing through this line can be described by the linear combination $P_1+\lambda P_2=0$.

If a point lies on both $P_1$ and $P_2$, it satisfies both equations, making the sum zero regardless of the value of $\lambda$. We are essentially creating a 'family' of infinite planes, and our goal is to find the specific member of this family that satisfies our condition.

Every plane has a unique signature: its normal vector. When we write our family equation as $(x+2y+3z+1)+\lambda (x-y-z-6)=0$, we group the terms to reveal the normal vector ${\vec{n}}_1$.

By collecting the coefficients, we obtain:

The normal vector is ${\vec{n}}_1=(1+\lambda )\hat{i}+(2-\lambda )\hat{j}+(3-\lambda )\hat{k}$. This vector is the 'DNA' of our target plane.

The problem states that our plane is perpendicular to a third plane, $P_3:-2x+y+z+8=0$. The normal vector of this third plane is ${\vec{n}}_2=-2\hat{i}+1\hat{j}+1\hat{k}$.

If two planes are perpendicular, their normal vectors must also be perpendicular. We test for this using the dot product: ${\vec{n}}_1\cdot {\vec{n}}_2=0$.

This leads us to the following equation:

Expanding the terms, we get $-2-2\lambda +2-\lambda +3-\lambda =0$. The constants simplify, leaving us with $-4\lambda +3=0$, which yields:

Substituting $\lambda =\frac{3}{4}$ back into our family equation $(x+2y+3z+1)+\frac{3}{4}(x-y-z-6)=0$, we multiply by 4 to clear the fraction:

The final equation of the plane is:

By plugging in the point $(0,1,1)$, we see that $7(0)+5(1)+9(1)=14$. It fits perfectly. You have successfully navigated the intersection of planes, mastered the normal vector, and applied the dot product to lock in the solution.