If the position vectors of the points A, B, C, D given by $\hat{i}+2\hat{j}+3\hat{k}$ , $2\hat{i}-\hat{j}+2\hat{k}$,
$\frac{1}{4}(7 \hat{i}+15\hat{j}+15 \hat{k})$ and $\frac{1}{3}[7\hat{i}+2\hat{j}+(5+3a)\hat{k}]$ respectively are such that |AC| =|BD| , then $16(3a-1)^{2}$=