If the line, \(\frac{{x - 1}}{2} = \frac{{y + 1}}{3} = \frac{{z - 2}}{4}{\rm{\;}}\) meets the plane, \(x + 2y + 3z = 15{\rm{\;}}\) at a point P, then the distance of P from the origin is:
1
\(\frac{{\sqrt 5 }}{2}\)
2
\(2\sqrt 5\)
3
\(\frac{9}{2}\)
4
\(\frac{7}{2}\)