The distance of the point $(-1,-5,-10)$ from the point of intersection of the line…
The distance of the point $(-1,-5,-10)$ from the point of intersection of the line $\frac{x-2}{3}=\frac{y+1}{4}=\frac{z-2}{12}$ and the plane $x-y+z=5$ is
13 units.
12 units.
5 units.
16 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
$\mathrm{P} \equiv(3 \lambda+2,4 \lambda-1,12 \lambda+2)$
This point lies on the plane
$\begin{aligned}
& x-y+z=5 \\
& \therefore \quad 3 \lambda+2-(4 \lambda-1)+12 \lambda+2=5 \\
& \Rightarrow 11 \lambda=0 \\
& \Rightarrow \lambda=0 \\
& \therefore \quad \mathrm{P} \equiv(2,-1,2) \\
& \text { Let } \mathrm{Q} \equiv(-1,-5,-10) \\
& \therefore \quad \mathrm{PQ}=\sqrt{(-1-2)^2+(-5+1)^2+(-10-2)^2} \\
& =\sqrt{9+16+144} \\
& =13 \text { units } \\
&
\end{aligned}$