The square of the distance of the point \(\displaystyle \left(\frac{15}{7},\frac{32}{7},7\right) \) from the line \(\displaystyle \frac{x+1}{3}=\frac{y+3}{5}=\frac{z+5}{7} \) in direction of the vector \( \hat{i}+4\hat{j}+7\hat{k} \) is:
\( 44 \)
\( 41 \)
\( 54 \)
\( 66 \)
