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(l, –9, 2) from the line L is
The line \(L\) passes through the point \( P(0, 1, 2) \), and intersects the given line and the plane. The direction vector for the line \(L\) is calculated by solving the system of equations using the direction of the given line and the plane. The distance formula for a point to a line is used, yielding:
\[
PQ = \sqrt{16 + 49 + 9} = \sqrt{74}
\]