1. Write the equations of the lines

• Line $L_1$ passes through $(1,2,3)$ and is parallel to the $z$-axis. So its direction vector is $\vec d_1=(0,0,1)$.

  Parametric form: $L_1:(x,y,z)=(1,2,3)+t(0,0,1)$ Hence, $x=1,\quad y=2,\quad z=3+t$

• Line $L_2$ passes through $(\lambda,5,6)$ and is parallel to the $y$-axis. So its direction vector is $\vec d_2=(0,1,0)$.

  Parametric form: $L_2:(x,y,z)=(\lambda,5,6)+s(0,1,0)$ Hence, $x=\lambda,\quad y=5+s,\quad z=6$

2. Find the shortest distance between $L_1$ and $L_2$

For skew lines, $\text{Shortest distance} = \frac{|(\vec r_2-\vec r_1)\cdot(\vec d_1\times \vec d_2)|}{|\vec d_1\times \vec d_2|}$

Take $\vec r_1=(1,2,3),\quad \vec r_2=(\lambda,5,6)$ So, $\vec r_2-\vec r_1=(\lambda-1,3,3)$

Now, $\vec d_1\times \vec d_2=(0,0,1)\times(0,1,0)=(-1,0,0)$ Thus, $|\vec d_1\times \vec d_2|=1$

Therefore, $\text{distance}=|((\lambda-1,3,3)\cdot(-1,0,0))|=|-(\lambda-1)|=|\lambda-1|$

Given shortest distance is $3$, so $|\lambda-1|=3$

Hence, $\lambda=4 \quad \text{or} \quad \lambda=-2$

Since $\lambda_2<\lambda_1$, $\lambda_1=4,\quad \lambda_2=-2$

3. Coordinates of the point

The point is $(\lambda_1,\lambda_2,7)=(4,-2,7)$

4. Find the distance of $(4,-2,7)$ from line $L_1$

Line $L_1$ is given by $x=1,\quad y=2$ with $z$ arbitrary.

So the perpendicular distance from a point $(x_0,y_0,z_0)$ to this line is just the distance in the $xy$-plane from $(x_0,y_0)$ to $(1,2)$: $d^2=(x_0-1)^2+(y_0-2)^2$

For $(4,-2,7)$, $d^2=(4-1)^2+(-2-2)^2=3^2+(-4)^2=9+16=25$

5. Check options

The required square of the distance is $\boxed{25}$ So the correct option is A.