The length of perpendicular drawn from the point ( 2, 3, 4 ) to...

Let $P$ be the foot of the perpendicular drawn from $A ( 2, 3, 4 )$ to the given line $l$.

Then, $\dfrac{x - 4}{- 2} = \dfrac{y}{6} = \dfrac{z - 1}{- 3}$

Now, any point on the line $l$ is given by

$x = 4 - 2\lambda, y = 6\lambda, z = 1 - 3\lambda$

The coordinates of $P$ are $( 4 - 2\lambda, 6\lambda, 1 - 3\lambda )$

The direction ratios of $AP$ are

$( 4 - 2\lambda - 2, 6\lambda - 3, 1 - 3\lambda - 4 )$

i.e. $( 2 - 2\lambda, 6\lambda - 3, - 3 - 3\lambda )$

And the direction ratios of $l$ are$- 2, 6$ and $- 3$.

Given, $AP ⊥ l$

$∴ - 2 ( 2 - 2\lambda ) + 6 ( 6\lambda - 3 ) - 3 ( - 3 - 3\lambda ) = 0$

$⇒ \lambda = \dfrac{13}{49}$

$∴ AP^{2} = ( 4 - 2\lambda - 2{)}^{2} + ( 6\lambda - 3{)}^{2} + ( 1 - 3\lambda - 4{)}^{2}$

$= 22 - 26\lambda + 49{λ}^{2}$

Put $\lambda = \dfrac{13}{49}$, we get

$AP^{2} = \dfrac{909}{49} ⇒ AP = \dfrac{3}{7}\sqrt{101}$