If $\overrightarrow{a} and \overrightarrow{b}$ are non zero non collinear vectors, then $\left[\overrightarrow{a} \overrightarrow{b} \overrightarrow{i}\right] \hat{i} +\left[\overrightarrow{a} \overrightarrow{b} \overrightarrow{j}\right] \hat{j}+ \left[\overrightarrow{a} \overrightarrow{b} \overrightarrow{k}\right] \hat{k}$ is equal to

Let $\overrightarrow{a}x\overrightarrow{b}=x\hat{i}+y\hat{j}+z\hat{k}$. Therefore, 

$\left[\overrightarrow{a}\overrightarrow{b}\hat{i}\right]=\left(\overrightarrow{a}\times \overrightarrow{b}\right)\hat{i}=x$

$\left[\overrightarrow{a}\overrightarrow{b}\hat{j}\right]=\left(\overrightarrow{a}\times \overrightarrow{b}\right)\hat{j}=y$

$\left[\overrightarrow{a}\overrightarrow{b}\hat{k}\right]=\left(\overrightarrow{a}\times \overrightarrow{b}\right)\hat{k}=z$

Hence,  $\left[\overrightarrow{a}\overrightarrow{b}\hat{i}\right]\hat{i} + \left[\overrightarrow{a}\overrightarrow{b}\hat{j}\right]\hat{j} =\left[\overrightarrow{a}\overrightarrow{b}\hat{k}\right]\hat{k} = x\hat{i} + y\hat{j} + z\hat{k} = \overrightarrow{a }\times \overrightarrow{b}$