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