The vectors $\bar{a}, \bar{b}$ and $\bar{c}$ are such that $|\overline{\mathrm{a}}|=2,|\overline{\mathrm{~b}}|=4,|\overline{\mathrm{c}}|=4$. If the projection of $\overline{\mathrm{b}}$ on $\overline{\mathrm{a}}$ is equal to projection of $\overline{\mathrm{c}}$ on $\overline{\mathrm{a}}$ and $\overline{\mathrm{b}}$ is perpendicular to $\overline{\mathrm{c}}$, then the value of $|\overline{\mathrm{a}}+\overline{\mathrm{b}}-\overline{\mathrm{c}}|$ is

Two tangents to the circle $x^2+y^2=4$ at the points A and B meet at $\mathrm{P}(-4,0)$. Then the area of quadrilateral PAOB, where ' $O$ ' is the origin is

The lines $\frac{6 x-6}{18}=\frac{y+1}{3}=\frac{z-1}{5} \quad$ and $\frac{3 x+6}{12}=\frac{y-1}{3}=\frac{z+1}{2}$ are $\ldots$

$$ \int x^2 \cos x d x= $$