Given that $\vec{a}$ is a perpendicular to $\vec{b}$ and p is a non-zero scalar, then a vector $\vec{r}$ satisfying $p \vec{r}+(\vec{r} . \vec{b}) \vec{a} = \vec{c}$ is given by

$\bar{r}=\frac{\bar{c}}{p}-\frac{\bar{c} . \bar{b}}{p^2} \bar{a}$

We have $p \overline{r}+(\overline{r} . \overline{b}) \overline{a}=\overline{c}$. Taking dot by vector $\vec{b}$, we get

$p \overline{r} . \overline{b}+(\overline{r} . \overline{b}) \overline{a} . \overline{b}=\overline{c} . \overline{b} \Rightarrow p \overline{r} . \overline{b}+0=\overline{c} . \overline{b} \Rightarrow \bar{r} . \bar{b}=\frac{\bar{c} . \bar{b}}{p}$

$\Rightarrow p \overline{r}+\frac{\overline{c} . \overline{b}}{p} \overline{a}=\overline{c} \Rightarrow \overline{r}=\frac{\overline{c}}{p}-\frac{\overline{c} . \overline{b}}{p^2} \overline{a}$

Hence (1) is correct answer.