If $\vec{a} \times \vec{b}=\vec{c}, \vec{b} \times \vec{c}=\vec{a}$ where $\vec{c} \neq \vec{0}$, then:

$|\vec{a}|=|\vec{c}|,|\vec{b}|=1$

$\vec{a} \times \vec{b}=\vec{c}, \vec{b} \times \vec{c}=\vec{a}$

Taking cross with $\vec{b}$ in first equation, we get

$\vec{b} \times(\vec{a} \times \vec{b})=\vec{b} \times \vec{c}=\vec{a}$

$\Rightarrow|\vec{b}|^2 \vec{a}-(\vec{a} . \vec{b}) \vec{b}=\vec{a}$

$\Rightarrow|\vec{b}|^2=1$  and  $\vec{a} . \vec{b}=0$

Also, $|\vec{a} \times \vec{b}|=|\vec{c}|$

$\Rightarrow|\vec{a}||\vec{b}| \sin \frac{\pi}{2}=|\vec{c}|$

$\Rightarrow|\vec{a}|=|\vec{c}|$

Hence (1) is correct answer.