### Three Dimensional Geometry Question 29

#### Question 29 - 31 January - Shift 2

Let the plane $P: 8 x+\alpha_1 y+\alpha_2 z+12=0$ be parallel to the line $L: \frac{x+2}{2}=\frac{y-3}{3}=\frac{z+4}{5}$. If the intercept of $P$ on the $y$-axis is 1 , then the distance between $P$ and $L$ is :

(1) $\sqrt{14}$

(2) $\frac{6}{\sqrt{14}}$

(3) $\sqrt{\frac{2}{7}}$

(4) $\sqrt{\frac{7}{2}}$

## Show Answer

#### Answer: (1)

#### Solution:

#### Formula: Equation Of A Plane

P: $8 x+\alpha_1 y+\alpha_2 z+12=0$

$L: \frac{x+2}{2}=\frac{y-3}{3}=\frac{z+4}{5}$

$\because P$ is parallel to $L$

$\Rightarrow 8(2)+\alpha_1(3)+5(\alpha_2)=0$

$\Rightarrow 3 \alpha_1+5(\alpha_2)=-16$

Also y-intercept of plane $P$ is 1

$\Rightarrow \alpha_1=-12$

And $\alpha_2=4$

$\Rightarrow$ Equation of plane $P$ is $2 x-3 y+z+3=0$

$\Rightarrow$ Distance of line $L$ from Plane $P$ is

$=|\frac{0-3(6)+1+3}{\sqrt{4+9+1}}|$

$=\sqrt{14}$