Let the line $L$ pass through the point $(0, 1, 2),$ intersect the line $\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}$ and be parallel to the plane $2x+y-3z=4$. Then the distance of the point $P(1, –9, 2)$ from the line $L$ is

Given,

The line $L$ pass through the point $(0, 1, 2),$ intersect the line $\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}$ and be parallel to the plane $2x+y-3z=4$,

Now plotting the diagram of above data we have,

Now let the direction of line which passes through $\left(0,1,2\right)$ be $\left(a,b,c\right)$

Now the line $L \& L_1$ are coplanar,

So coplanar condition we get,

$\left|\begin{array}{ccc}a& b& c\\ 2-1& 3-2& 4-3\\ 2& 3& 4\end{array}\right|=0$

$\Rightarrow \left|\begin{array}{ccc}a& b& c\\ 1& 1& 1\\ 2& 3& 4\end{array}\right|=0$

$\Rightarrow a-2b+c=0 .......\left(1\right)$

And the line $L$ is perpendicular to the normal vector of the plane, 

So by perpendicular condition we get,
$2a+b-3c=0 ......\left(2\right)$

So, from equation $\left(1\right) \& \left(2\right)$ we get,

$a=b=c$

So, equation of line $L$ will be,

$\Rightarrow L=\frac{x}{1}=\frac{y-1}{1}=\frac{z-2}{1}=\lambda$

So any point on $L$ can be taken as $A\left(\lambda , 1+\lambda , 2+\lambda \right)$,

Now $\overrightarrow{AP}$ will be perpendicular to the direction ratio of the line $L$,

So, again by perpendicular condition we get,

$\overrightarrow{AP}·\left(\hat{i}+\hat{j}+\hat{k}\right)=0$

$\Rightarrow \lambda -1+\lambda +10+\lambda =0$

$\Rightarrow 3\lambda +9=0$

$\Rightarrow \lambda =-3$

Hence,$A\left(-3, -2, -1\right) \& P\left(1, -9, 2\right)$

So, by distance formula we get, $AP=\sqrt{74}$