Consider the following statements

p : the switch $\mathrm{S}_1$ is closed.

q : the switch $\mathrm{S}_2$ is closed.

$r$ : the switch $\mathrm{S}_3$ is closed.

Then the switching circuit represented by the statement $(p \wedge q) \vee(\sim p \wedge(\sim q \vee p \vee r))$ is

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