In triangle ABC , the point P divides BC internally in the ratio $3: 4$ and Q divides CA internally in the ratio $5: 3$. If AP and BQ intersect in a point $G$, then $G$ divides $A P$ internally in the ratio

Let $\bar{u}, \bar{v}, \bar{w}$ be the vectors such that $|\overline{\mathrm{u}}|=1,|\overline{\mathrm{v}}|=2,|\overline{\mathrm{w}}|=3$. If the projection $\overline{\mathrm{v}}$ along $\overline{\mathrm{u}}$ is equal to that of $\overline{\mathrm{w}}$ along $\overline{\mathrm{u}}$ and the vectors $\overline{\mathrm{v}}, \overline{\mathrm{w}}$ are perpendicular to each other then $|\overline{\mathrm{u}}-\overline{\mathrm{v}}+\overline{\mathrm{w}}|$ equals

The projection of the line segment joining the points $(2,1,-3)$ and $(-1,0,2)$ on the line whose direction ratios are $3,2,6$ is

If $\bar{a}, \bar{b}, \bar{c}$ are three vectors such that $|\bar{a}|=3$, $|\bar{b}|=5,|\bar{c}|=7$ then $|\bar{a}-\bar{b}|^2+|\bar{b}-\bar{c}|^2+|\bar{c}-\bar{a}|^2$ does not exceed