The square of the distance of the point $\left( \frac { 15 } { 7 } , \frac { 32 } { 7 } , 7 \right)$ from the line $\frac { x + 1 } { 3 } = \frac { y + 3 } { 5 } = \frac { z + 5 } { 7 }$ in the direction of the vector $\hat { i } + 4 \hat { j } + 7 \hat { k }$ is :\\
(1) 54\\
(2) 44\\
(3) 41\\
(4) 66