The vertices B and C of a ∆ABC lie on the line \(\frac{{x + 2}}{3} = \frac{{y - 1}}{0} = \frac{z}{4}\) such that BC = 5 units. Then, the area (in sq. units) of this triangle, given that the point A (1, -1, 2) is:

• A. \(5\sqrt {17} \)
• B. \({\rm{\;}}2\sqrt {34} \)
• C. 6
• D. \(\sqrt {34} \)