The length of perpendicular drawn from the point $(2,3,4)$ to line $\frac{4-x}{2}=\frac{y}{6}=\frac{1-z}{3}$, is

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

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

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 $\mathrm{AP}$ 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, $AP\perp l$

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

$\Rightarrow \lambda =\frac{13}{49}$

$\therefore A{P}^2=(4-2\lambda -2{)}^2+(6\lambda -3{)}^2+(1-3\lambda -4{)}^2$

$=22-26\lambda +49{\lambda }^2$

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

$A{P}^2=\frac{909}{49}\Rightarrow AP=\frac{3}{7}\sqrt{101}$