Let $\vec{a}$ and $\vec{b}$ be unit vectors that are perpendicular to each other, then $[\vec{a}+(\vec{a} \times \vec{b}), \vec{b}+(\vec{a} \times \vec{b}), \vec{a} \times \vec{b}]$ will always be equal to:

1

$[\vec{a}+(\vec{a} \times \vec{b}), \vec{b}+(\vec{a} \times \vec{b}), \vec{a} \times \vec{b}]$

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

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

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

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

= 1 (as $\vec{a} . \vec{b}=0, \vec{a} . (\vec{a} \times \vec{b})= 0$)

Hence (1) is correct answer.