Let $\bar{a}, \bar{b}, \bar{c}$ be three non-coplanar vectors and $\overline{\mathrm{p}}, \overline{\mathrm{q}}, \overline{\mathrm{r}}$ defined by the relations

$$\overline{\mathrm{p}}=\frac{\overline{\mathrm{b}} \times \overline{\mathrm{c}}}{[\overline{\mathrm{a}} \overline{\mathrm{~b}} \overline{\mathrm{c}}]}, \overline{\mathrm{q}}=\frac{\overline{\mathrm{c}} \times \overline{\mathrm{a}}}{[\overline{\mathrm{a}} \overline{\mathrm{~b}} \overline{\mathrm{c}}]}, \overline{\mathrm{r}}=\frac{\overline{\mathrm{a}} \times \overline{\mathrm{b}}}{[\overline{\mathrm{a}} \overline{\mathrm{~b}} \overline{\mathrm{c}}]}$$

then the value of the expression $(\overline{\mathrm{a}}+\overline{\mathrm{b}}) \cdot \overline{\mathrm{p}}+(\overline{\mathrm{b}}+\overline{\mathrm{c}}) \cdot \overline{\mathrm{q}}+(\overline{\mathrm{c}}+\overline{\mathrm{a}}) \cdot \overline{\mathrm{r}}$ is equal to

If $A=\left[\begin{array}{cc}2 & -2 \\ 4 & 3\end{array}\right]$, then $A^{-1}=$

The differential equation of family of circles, whose centres are on the X -axis and also touch the Y -axis is

A ball ' $A$ ' is projected vertically upwards with certain initial speed. Another ball 'B' of same mass is projected at an angle of $30^{\circ}$ with vertical with the same initial speed. At the highest point, the ratio of potential energy of ball A to that of ball B will be

$$\left(\sin 90^{\circ}=1, \sin 60^{\circ}=\cos 30^{\circ}=\frac{\sqrt{3}}{2}, \sin 30^{\circ}=\cos 60^{\circ}=\frac{1}{2}\right)$$