The Cartesian equation of a line is $2 x-2=3 y+1=6 z-2$, then the vector equation of the line is

Let $\overline{\mathrm{a}}=2 \hat{\mathrm{i}}+\hat{\mathrm{j}}-2 \hat{\mathrm{k}}$ and $\overline{\mathrm{b}}=\hat{\mathrm{i}}+\hat{\mathrm{j}}$. Let $\overline{\mathrm{c}}$ be a vector such that $|\overline{\mathrm{c}}-\overline{\mathrm{a}}|=3$ and $|(\overline{\mathrm{a}} \times \overline{\mathrm{b}}) \times \overline{\mathrm{c}}|=3$ and the angle between $\bar{c}$ and $\bar{a} \times \bar{b}$ is $30^{\circ}$, then $\bar{a} \cdot \bar{c}$ is equal to

If $\int\left(\frac{4 e^x-25}{2 e^x-5}\right) d x=A x+B \log \left(2 e^x-5\right)+c \quad$ (where c is a constant of integration) then

The converse of $[p \wedge(\sim q)] \rightarrow r$ is