OABC is a tetrahedron of volume V and $\angle AOB={45}^0.$ If $l$ denotes the length of altitude drawn from C to face OAB, $BC=\sqrt{2},V=\frac{2}{3},OA+OB=4,$ then $l^2+O{C}^2$ is equal to 

$Let\theta =BC\wedge$ (normal to plane OAB)

$V=\frac{2}{3}=\frac{1}{3}\left(ar\left(\Delta OAB\right)\right).BC\cos \theta =\frac{1}{6}.OA.OB\frac{1}{\sqrt{2}}\sqrt{2}\cos \theta$

$\frac{2}{3}=\frac{OA.OB.\cos \theta }{6}\therefore OA.OB=4\sec \theta \ge 4$

$ButOA+OB=4\Rightarrow OA.OB\le 4\Rightarrow \cos \theta =1\&OA=OB$

$\therefore l=\sqrt{2}andOC=\sqrt{O{B}^2+B{C}^2}=\sqrt{4+2}=\sqrt{6}$