Let $\hat{a}$ and $\hat{b}$ be unit vector that are mutually perpendicular, then for any arbitrary $\vec{r}$ :  

$\vec{r}=(\vec{r} . \hat{a}) \hat{a}+(\vec{r} . \hat{b}) \hat{b}+(\vec{r} .(\hat{a} \times \hat{b}))(\hat{a} \times \hat{b})$

Let  $\vec{r}=x_1 \hat{a}+x_2 \hat{b}+x_3(\hat{a} \times \hat{b})$

$\Rightarrow \vec{r} . \hat{a}=x_1+x_2 \hat{a} . \hat{b}+x_3 \hat{a} .(\hat{a} \times \hat{b})=x_1$

Also,  $\vec{r} . \hat{b}=x_1 \hat{a} . \hat{b}+x_2+x_3 \hat{b} .(\hat{a} \times \hat{b})=x_2$

and  $\vec{r} .(\hat{a} \times \hat{b})=x_1 \hat{a} .(\hat{a} \times \hat{b})+x_2 \hat{b} .(\hat{a} \times \hat{b})+x_3(\hat{a} \times \hat{b}) .(\hat{a} \times \hat{b})=x_3$

$\Rightarrow \vec{r}=(\vec{r} . \hat{a}) \hat{a}+(\vec{r} . \hat{b}) \hat{b}+(\vec{r} .(\hat{a} \times \hat{b}))(\hat{a} \times \hat{b})$

Hence (1) is correct answer.