Let \(\vec e=\vec a\times\vec b\)
\(|\vec e|=|\vec a||\vec b|sin \alpha\)....(1)
(∵ α is angle between \(\vec a\) and \(\vec b\))
Then
\(=|\vec e||\vec c|cos\theta\)
(∵θ is angle between \(\vec e\)and \(\vec c\) ⇒ θ is angle between \(\vec a\times\vec b\) and \(\vec c\))
\(\alpha=\cfrac{\pi}2\implies\vec a\) and \(\vec b\) are perpendicular.
Also \(\vec e\) is perpendicular to both \(\vec a\) and \(\vec b\).
θ = 0 ⇒ \(\vec c\)is perpendicular to both \(\vec a\) and \(\vec b\).
\(\therefore\vec a,\,\vec b,\,\vec c\) are mutually perpendicular.
\(\therefore\vec a.\vec b=\vec b.\vec c=\vec c.\vec a=0\)