Triangle ∆abc has area of s m2. points n and p are taken on rays bc and ca respectively so that the length bn = 2bc and cp = 3ca. find the area of ∆pbn. (hints: find the area of ∆pbc in terms of s by looking for triangles with equal heights. then compare the area of ∆pbc to the area of ∆pnc.)

Triangle PBC has the same height as triangle ABC, but the base, PC, is 3 times longer, so the area of the triangle PBC is 3 times the area of the triangle ABC. Since triangle ABC has area S, triangle PBC has area 3S.
Triangle PCN has the same base and height as triangle PBC, so it has the same area. The area of triangle PCN is 3S.
Triangle PBN is made up of triangles PBC and PCN, so the area of triangle PBN is 3S + 3S = 6S m^2.

I think it’s the 2nd one sorry if I’m wrong