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Q.
The length of the perpendicular drawn from $(1,2,3)$ to the line $\frac{x-6}{3}=\frac{y-7}{2}=\frac{z-7}{-2}$ is
Three Dimensional Geometry
Solution:
Let $P$ be the point $(1, 2, 3)$ and $PN$ be the length of the perpendicular from $P$ on the given line.
Coordinates of point $N$ are $\left(3\lambda+6, 2\lambda+7,-2\lambda+7\right)$
Now $PN$ is perpendicular to the given line or vector
$3\vec{i}+2\vec{j}-2\vec{k}.$ Thus,
$3\left(3\lambda+6-1\right)+2\left(2\lambda+7-2\right)-2\left(-2+7-3\right)=0 $
or $ \lambda=-1$ Then, point $N$ is $(3, 5, 9)$
$\Rightarrow PN=7$