Since no-one else has replied I'll suggest an approach:

1. Go to parametric equation for the ellipse: x = 4 cos t, y = 3 sin t. Then the vertices of the triangle are A(0, 3), B(4, 0) and C(4 cos t, 3 sin t).

2. The area of a triangle whose vertices have coordinates (x1, y1), (x2, y2) and (x3, y3) is A = 1/2 |(x1 - x3)(y1- y3) - (y1 - y2)(x1 - x3)| (you should prove this - only takes a couple of lines).

3. Sub in the coordinates of the vertices of the triangle and solve A = 3. You get a trig equation that you may need help solving.

There's probably a much more elegant approach - maybe it's no coincidence that if you consider the point D(0, -3) then triangle ABD has an area of 12 = 4 x 3 .....