Consider a tetrahedron ABCD such that the areas of the triangles ABC, BCD, ACD, ABD are $k,l,m,n$ respectively. The volume of tetrahedron is $(1/6)$ cubic units . If each face ACB, ACD, BCD of the tetrahedron subtends a right angle at C, then we have to find the geometric mean of $k,l,m$.

I remember that $\frac{1}{3}(base)(height)=\frac{1}{6}$.

in this base area is know , I am confused how to find height