Question

The distance of the 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. $2\sqrt{14}$
• B. 8
• C. $3\sqrt{21}$
• D. 13

Answer 1

The correct option is D

13

Given equation of line is
$\frac{x-2}{3}=\frac{y+1}{4}=\frac{z-2}{12}=\lambda \left[say\right]...\left(i\right)$
and equation of plane is
$x-y+z=16.........\left(ii\right)$
Any point on the line (i) is, $(3\lambda +2,4\lambda -1,12\lambda +2)$
Let this point of intersection of the line and plane.
$\therefore (3\lambda +2)-(4\lambda -1)+(12\lambda +2)=16\Rightarrow 11\lambda +5=16\Rightarrow 11\lambda =11\Rightarrow \lambda =1$
So, the point of intersection is (5, 3, 14).
Now, distance between the points (1, 0, 2) and (5, 3, 14)
$=\sqrt{(5-1{)}^2+(3-0{)}^2+(14-2{)}^2}$
$=\sqrt{16+9+144}=\sqrt{169}=13$