Question

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 :

• A. $\sqrt{\frac{2}{7}}$
• B. $\frac{6}{\sqrt{14}}$
• C. $\sqrt{\frac{7}{2}}$
• D. $\sqrt{14}$

Answer 1

Answer: (D)

$\sqrt{14}$

Answer 2

P: 8x+α1​y+α2​z+12=0
L:2x+2​=3y−3​=5z+4​
∵P is parallel to L
⇒8(2)+α1​(3)+5(α2​)=0
⇒3α1​+5(α2​)=−16
Also y-intercept of plane P is 1
⇒α1​=−12
And α2​=4
⇒ Equation of plane P is 2x−3y+z+3=0
⇒ Distance of line L from Plane P is
=∣∣​4+9+1​0−3(6)+1+3​∣∣​
=14​
So, the correct option is (D) : $\sqrt{14}$