If $$\mathrm{A}$$ and $$\mathrm{B}$$ are the foot of the perpendicular drawn from the point $$\mathrm{Q}(\mathrm{a}, \mathrm{b}, \mathrm{c})$$ to the planes $$\mathrm{YZ}$$ and $$\mathrm{ZX}$$ respectively, then the equation of the plane through the points $$\mathrm{A}, \mathrm{B}$$, and $$\mathrm{O}$$ is (where $$\mathrm{O}$$ is the origin)

If $$\mathrm{A}=(-2,2,3), \mathrm{B}=(3,2,2), \mathrm{C}=(4,-3,5)$$ and $$\mathrm{D}=(7,-5,-1)$$ Then the projection of $$\overline{\mathrm{AB}}$$ on $$\overline{\mathrm{CD}}$$ is

The Cartesian equation of a plane which passes through the points $$\mathrm{A}(2,2,2)$$ and making equal nonzero intercepts on the co-ordinate axes is

The co-ordinates of the foot of the perpendicular drawn from the point $$2 \hat{i}-\hat{j}+5 \hat{k}$$ to the line $$\vec{r}=(11 \hat{i}-2 \hat{j}-8 \hat{k})+\lambda(10 \hat{i}-4 \hat{j}-11 \hat{k})$$ are