$\hat i\times (\hat j+\hat k)=\hat i\times \hat j+\hat i\times \hat k$
$\qquad=\hat k-\hat j$
Similarly $\hat j\times (\hat k+\hat i)=\hat i-\hat k$
and
$\hat k\times(\hat i+\hat j)=\hat j-\hat i$
Adding all the three we get
$\hat i\times (\hat j+\hat k)+\hat j\times (\hat k+\hat i)+\hat k\times (\hat i+\hat j)$
$=\hat k-\hat j+\hat i-\hat k+\hat j-\hat i=\overrightarrow 0$