Question

The distance of point $(1,0,2)$ from the point of intersection of the line $\frac{x-2}{3}=\frac{y+1}{4}=\frac{z-2}{12}$ and the plane $x-y+z=16$ is

• A. $3\sqrt{21}$
• B. $2\sqrt{14}$
• C. $13$
• D. $8$

Answer 1

The correct option is C $13$

$\frac{x-2}{3}=\frac{y+1}{4}=\frac{z-2}{12}=\lambda$

$x=3\lambda +2$ $y=4\lambda -1$ $z=12\lambda +2$

$(x,y,z)\to x-y+z=16$ should satisfy

$3\lambda +2-(4\lambda -1)+12\lambda +2=16$

$3\lambda +2-4\lambda +1+12\lambda +2=16$

$11\lambda =11$

$\lambda =1$

$(x,y,z)=(5,3,14)$ - point of intersection of line

Distance b/w (1,0,2) & (5,3,14)

$d=\sqrt{(5-1{)}^2+(3-0{)}^2+(14-2{)}^2}$

$=\sqrt{16+9+144}$

$=\sqrt{169}$

$d=13$

$\therefore$ option C = 13