Let the line L pass through the point (0, 1, 2) , intersect the line $\frac{\mathrm{x}-1}{2}=\frac{\mathrm{y}-2}{3}=\frac{\mathrm{z}-3}{4}$ 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

$\frac{\mathrm{x}-1}{2}=\frac{\mathrm{y}-2}{3}=\frac{\mathrm{z}-3}{4}=\mathrm{r}$
Any point on the line is

Dr’s of $\overleftrightarrow{\mathrm{AB}}$ are
(2r + 1, 3r + 1, 4r + 1)
$\because \overrightarrow{\mathrm{AB}}{\parallel }^{\mathrm{le}}$ to the plane 2x + y - 3z = 4
$\begin{array}{l}\Rightarrow \mathrm{al}+\mathrm{bm}+\mathrm{cn}=0\\ \Rightarrow 2(2\mathrm{r}+1)+1(3\mathrm{r}+1)-3(4\mathrm{r}+1)=0\\ \Rightarrow \mathrm{r}=0\end{array}$
B(1, 2, 3)
$\begin{array}{l}{\mathrm{L}}^{\mathrm{\prime }}=\frac{\mathrm{x}-0}{1}=\frac{\mathrm{y}-1}{1}=\frac{\mathrm{z}-2}{1}=\mathrm{t}\\ \Rightarrow \mathrm{Q}(\mathrm{t},\mathrm{t}+1,\mathrm{t}+2);\mathrm{P}(1,-9,2)\end{array}$
Dr’s of $\overleftrightarrow{\mathrm{PQ}}$ = (t - 1)1 + (t + 10)1 + t = 0
$\begin{array}{l}\overline{\mathrm{PQ}}=(-4,7,-3)\mathrm{t}=-3\\ \left|\overline{\mathrm{PQ}}\right|=\sqrt{16+49+9}=\sqrt{74}\end{array}$