### Three Dimensional Geometry Question 20

#### Question 20 - 30 January - Shift 1

If the equation of the plane passing through the point $(1,1,2)$ and perpendicular to the line $x-3 y+2 z-1=04 x-y+z$ is $Ax+By+Cz=1$, then $140(C-B+A)$ is equal to ________

## Show Answer

#### Answer: 15

#### Solution:

#### Formula: Equation of a plane (6.7), Direction ratio and direction cosines

$ \begin{aligned} & x-3 y+2 z-1=0 \\ & 4 x-y+z=0 \\ & \vec{n} _1 \times \vec{n} _2= \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -3 & 2 \\ 4 & -1 & 1 \end{vmatrix} \\ & =-\hat{i}+7 \hat{j}+11 \hat{k} \end{aligned} $

$Dr^{s}$ of normal to the plane is $-1,7,11$

Equation of plane :

$ \begin{aligned} & -1(x-1)+7(y-1)+11(z-2)=0 \\ & -x+7 y+11 z=28 \\ & \frac{-1}{28} x+\frac{7 y}{28}+\frac{11 z}{28}=1 \end{aligned} $

$Ax+By+Cz=1$

$140(C-B+A)=140(\frac{11}{28}-\frac{7}{28}-\frac{1}{28})$

$=140 \times \frac{3}{28}=15$