Question

The distance of the point $\left(2,3,2\right)$ from the plane $3x+4y+4z=23$ measured parallel to the line $\frac{x+3}{1}=\frac{y−6}{−2}=\frac{z−1}{1}$ is

• A. $\sqrt{108}$ units
• B. $12$ units
• C. $\sqrt{54}$ units
• D. $\sqrt{236}$ units

Answer 1

Answer: (C)

Equation of the line passing through $\left(2,3,2\right)$ and parallel to the given line is $\frac{x−2}{1}=\frac{y−3}{−2}=\frac{z−2}{1}$
Any general point on this line is $\left(λ+2,−2λ+3,λ+2\right)$
This must satisfy the given equation of the plane
$⇒3λ+6−8λ+12+4λ+8=23$
$⇒−λ+26=23⇒λ=3$
The point on the plane is $\left(5,−3,5\right)$
Hence, the required distance $=\sqrt{9+36+9}=\sqrt{54}$
units