Question

The distance of the point (3, 8, 2) from the line $\frac{x-1}{2}=\frac{y-3}{4}=\frac{z-2}{3}$ measured parallel to the plane $3x+2y-2z+15=0$, is

• A. 2
• B. 3
• C. 6
• D. 7

Answer 1

The correct option is D 7

Equation of plane parallel to $3x+2y-2z+15=0$ is

$3x+2y-2z+d=0$

Let it passes through $(3,8,2)$

$\Rightarrow 3\left(3\right)+2\left(8\right)-2\left(2\right)+d=0$

$\Rightarrow d=-21$

$\Rightarrow 3x+2y-2x-21=0$............(i)

Equation of line is

$\frac{x-1}{2}=\frac{y-3}{4}=\frac{z-2}{3}=a$

$\Rightarrow x=2a+1,y=4a+3,z=3a+2$

Replacing the values in (i)

$3(2a+1)+2(4a+3)-2(3a+2)-21=0$

$6a+3+8a+6-6a-4-21=0$

$\Rightarrow a=2$

$\Rightarrow x=5,y=11,z=8$

So the distance of $(3,8,2)$ along $3x+2y-2z+15=0$ along given line

$=\sqrt{(5-3{)}^2+{(11-8)}^2+{(8-2)}^2}$

$=\sqrt{4+9+36}=7$

So, option D is correct.