If the line \(\frac{x-1}{2}=\frac{y+1}{3}=\frac{z-2}{4}\) meets the plane \(x+2 y+3 z=15\) at the point \(\mathrm{P}\), then the distance of \(\mathrm{P}\) from the origin is

Let \(\frac{x-1}{2}=\frac{y+1}{3}=\frac{z-2}{4}=\mathrm{k}\) (say)
Let \(\mathrm{P}\) be the any point on the above line.
\(\therefore \quad \mathrm{P}=(2 \mathrm{k}+1,3 \mathrm{k}-1,4 \mathrm{k}+2)\)
The point \(\mathrm{P}\) lies on the plane
\(\begin{array}{ll}
\therefore & 2 \mathrm{k}+1+2(3 \mathrm{k}-1)+3(4 \mathrm{k}+2)=15 \\
\therefore & 2 \mathrm{k}+1+6 \mathrm{k}-2+12 \mathrm{k}+6=15 \\
\therefore & 20 \mathrm{k}=10 \\
\therefore & \mathrm{k}=\frac{1}{2} \\
\therefore & \mathrm{P}=\left(2, \frac{1}{2}, 4\right)
\end{array}\)
Distance of \(\mathrm{P}\) from origin is
\(\sqrt{2^2+\left(\frac{1}{2}\right)^2+4^2}=\sqrt{\frac{81}{4}}=\frac{9}{2}\)