Let $\frac{x-1}{2}=\frac{y+1}{3}=\frac{z-2}{4}=\lambda$

Let, coordinate of any point on line is $P=\left(2\lambda +1, 3\lambda -1, 4\lambda +2\right)$

If $P$ is intersection point of plane & line then $P$ will lie on plane $x+2y+3z=15$

$\therefore \left(2\lambda +1\right)+2\left(3\lambda -1\right)+3\left(4\lambda +2\right)=15$

$\Rightarrow \lambda =\frac{1}{2}$
$\therefore P=\left(2\times \frac{1}{2}+1, 3\times \frac{1}{2}-1, 4\times \frac{1}{2}+2\right)=\left(2, \frac{1}{2}, 4\right)$

The distance of a point $\left(x_1, y_1, z_1\right)$ from origin $O$ is $\sqrt{{\left(x_1\right)}^2+{\left(y_1\right)}^2+{\left(z_1\right)}^2}$ 

$\therefore OP=\sqrt{2^2+{\left(\frac{1}{2}\right)}^2+4^2}=\sqrt{\frac{81}{4}}=\frac{9}{2}$units.