Let $\overrightarrow{a}=x\hat{i}+y\hat{j}+z\hat{k}$, and using vector triple product, we have,
$\hat{i}\times \left(\overrightarrow{a}\times \hat{i}\right)=\left(\hat{i}.\hat{i}\right)\overrightarrow{a}-\left(\hat{i}.\overrightarrow{a}\right)\hat{i}=y\hat{j}+z\hat{k}$
Similarly,

$\hat{\mathit{j}}\mathit{\times }\left(\overrightarrow{\mathit{a}}\mathit{\times }\hat{\mathit{j}}\right)\mathit{=}x\hat{\mathit{i}}\mathit{+}z\hat{\mathit{k}}$

and $\hat{k}\mathit{\times }\left(\overrightarrow{a}\mathit{\times }\hat{k}\right)\mathit{=}x\hat{i}\mathit{+}y\hat{k}$

Thus, 

${\left|\hat{i}\times \left(\overrightarrow{a}\times \hat{i}\right)\right|}^2+{\left|\hat{j}\times \left(\overrightarrow{a}\times \hat{j}\right)\right|}^2+{\left|\hat{k}\times \left(\overrightarrow{a}\times \hat{k}\right)\right|}^2$
$={\left|y\hat{j}+z\hat{k}\right|}^2+{\left|x\hat{i}+z\hat{k}\right|}^2+{\left|x\hat{i}+y\hat{j}\right|}^2$

$=2|\overrightarrow{a}{|}^2=2(9)=18$