Question

A point $P$ moves in the plane $\prod :2x-3y+6z-4=0$ such that the area of $△PAB$, where $A\equiv (2,2,1)$ and $B\equiv (-1,-4,-1)$ is $14\text{sq. units}$. If the plane perpendicular to the plane $\prod$ containing the locus of $P$ are $6x+ay+bz+d_1=0$ and $6x+ay+bz+d_2=0,d_1>d_2$, then the value of $\left(\frac{d_1-d_2+a+b}{3}\right)$ is

Answer 1

[Note: $A$ and $B$ lies in the plane $\prod$]

Area of $△PAB$
$14=\frac{1}{2}\left(AB\right)\text{height}\Rightarrow \frac{1}{2}\times \sqrt{3^2+6^2+2^2}\times \text{height}=14\Rightarrow \text{height}=4$
$\therefore$ Locus of point $P$ is the pair of straight line parallel to $AB$

Now, the normal of the required plane i.e.
$6\hat{i}+a\hat{j}+b\hat{k}$ is perendicular to $\overrightarrow{AB}$ as well as perpendicular to $\overrightarrow{n}$ of the given plane $\prod$.
So,
$\overrightarrow{AB}=-3\hat{i}-6\hat{j}-2\hat{k},\overrightarrow{n}=2\hat{i}-3\hat{j}+6\hat{k}-18-6a-2b=0\Rightarrow 9+3a+b=0\cdots \left(1\right)12-3a+6b=0\Rightarrow 4-a+2b=\cdots \left(2\right)$
Solving equations $\left(1\right)$ and $\left(2\right)$, we get
$a=-2,b=-3$
The equation of the planes become
$6x-2y-3z+d_1=06x-2y-3z+d_2=0$
Distance between them is $8\text{units}$, so
$\frac{|d_1-d_2|}{\sqrt{6^2+2^2+3^2}}=8\Rightarrow d_1-d_2=56(\because d_1>d_2)\therefore \left(\frac{d_1-d_2+a+b}{17}\right)=\frac{56-2-3}{3}=17$