The distance of the point $(1,6,2)$ from the point of intersection of the line…
The distance of the point $(1,6,2)$ 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
11 units
12 units
13 units
14 units
Solution
Let $\frac{x-2}{3}=\frac{y+1}{4}=\frac{z-2}{12}=\lambda$
$\therefore \quad$ The co-ordinates of any point on the line are
$P \equiv(3 \lambda+2,4 \lambda-1,12 \lambda+2)$
This point lies on the plane $x-y+z=16$
$\therefore \quad 3 \lambda+2-(4 \lambda-1)+12 \lambda+2=16$
$\begin{aligned} & \Rightarrow 11 \lambda=11 \\ & \Rightarrow \lambda=1\end{aligned}$
$\therefore \quad \mathrm{P} \equiv(5,3,14)$
Let $\mathrm{Q} \equiv(1,6,2)$
$\therefore \quad \mathrm{PQ}=\sqrt{(1-5)^2+(6-3)^2+(2-14)^2}$
$=\sqrt{16+9+144}=13$ units