Let $\vec{a}$ and $\vec{b}$ are two non collinear vector such that $|\vec{a}|=1$. The angle of a triangle whose two sides are represented by the vector $\sqrt{3}(\vec{a} \times \vec{b})$ and $\vec{b}-(\vec{a} . \vec{b}) \vec{a}$ are

$\frac{\pi}{2}, \frac{\pi}{3}, \frac{\pi}{6}$

Let $\vec{r}_1=\sqrt{3}(\vec{a} \times \vec{b}), \vec{r}_2=\vec{b}-(\vec{a} . \vec{b}) \vec{a}$, clearly $\vec{r}_1$ and $\vec{r}_2$ are mutually perpendicular as $\vec{r}_2$ is coplanar with $\vec{a}$ and $\vec{b}$ and $\vec{r}_1$ is at right angle to the plane of $\vec{a}$ and $\vec{b}$. And $\left|\vec{r}_1\right|=\sqrt{3}|\vec{a} \times \vec{b}| \Rightarrow\left|\vec{r}_1\right|^2=3(\vec{a} \times \vec{b}) .(\vec{a} \times \vec{b})$

$=3\left((\vec{a} . \vec{a})(\vec{b} . \vec{b})-(\vec{a} . \vec{b})^2\right)=3\left(|\vec{b}|^2-(\vec{a} . \vec{b})^2\right)$

Also, $\left|\vec{r}_2\right|=(\vec{b}-(\vec{a} . \vec{b}) \vec{a}) .(\vec{b}-(\vec{a} . \vec{b}) \vec{a})=|\vec{b}|^2-2(\vec{a} . \vec{b})^2+(\vec{a} . \vec{b})^2(\vec{a} . \vec{a})$

$=|\vec{b}|^2-(\vec{a} . \vec{b})^2 \Rightarrow \frac{\left|r_1\right|}{\left|r_2\right|}=\sqrt{3}$

Thus angles are $\frac{\pi}{2}, \frac{\pi}{3}, \frac{\pi}{6}$

Hence (2) is correct answer.