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# The length of the perpendicular drawn from the point $(3,-1,11)$ on the line $\large\frac{x}{2}=\frac{y-2}{3}=\frac{z-3}{4}$ is?

$\begin{array}{1 1} (a)\:\:\sqrt {29}\:\:\:\qquad\:\:(b)\:\:\sqrt {33}\:\:\:\qquad\:\:(c)\:\:\sqrt {53}\:\:\:\qquad\:\:(d)\:\:\sqrt {65}\end{array}$

Toolbox:
• $d.r.$ of the line joining $(x_1,y_1,z_1)\:\:and\:\: (x_2,y_2,z_2)$ is $(x_1-x_2,y_1-y_2,z_1-z_2)$
Given line: $\large\frac{x}{2}=\frac{y-2}{3}=\frac{z-3}{4}=\lambda$
Any point on this line can be given in terms of $\lambda$ as $(2\lambda,3\lambda+2,4\lambda+3)$
Since the foot of $\perp$ from $P(3,-1,11)$ on this line lies on the line,
let $Q(2\lambda,3\lambda+2,4\lambda+3)$ be the foot of $\perp$
$\therefore\:d.r.$ of $\perp$ $\overline {PQ}$ is $(2\lambda-3,3\lambda+3,4\lambda-8)$
Since $\overline {PQ}$ is $\perp$ to the line, $(2,3,4).(\overline{PQ})=0$
$\Rightarrow\:4\lambda-6+9\lambda+9+16\lambda-32=0$
$\Rightarrow\:29\lambda-29=0\:\:\:or\:\:\:\lambda=1$
$\therefore\:Q(2,5,7)$
$\therefore\:$ The length of $\perp =distance\: \overline{PQ}=\sqrt {1+36+16}=\sqrt {53}$