Let a line passing through the point $(4,1,0)$ intersect the line $\mathrm{L}_1: \frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}$ at the point $A(\alpha, \beta, \gamma)$ and the line $\mathrm{L}_2: x-6=y=-z+4$ at the point $B(a, b, c)$. Then $\left|\begin{array}{lll}1 & 0 & 1 \\ \alpha & \beta & \gamma \\ a & b & c\end{array}\right|$ is equal to

Line $L_1$ passes through the point $(1,2,3)$ and is parallel to $z$-axis. Line $L_2$ passes through the point $(\lambda, 5,6)$ and is parallel to $y$-axis. Let for $\lambda=\lambda_1, \lambda_2, \lambda_2<\lambda_1$, the shortest distance between the two lines be 3 . Then the square of the distance of the point $\left(\lambda_1, \lambda_2, 7\right)$ from the line $L_1$ is