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 $$\lambda = \lambda_1, \lambda_2$$, $$\lambda_2 \lt \lambda_1$$, the shortest distance between the two lines be 3. Then the square of the distance of the point $$(\lambda_1, \lambda_2, 7)$$ from the line $$L_1$$ is

Line $$L_1$$ passes through $$P_1(1,2,3)$$ and is parallel to the $$z$$-axis.
Hence a direction vector of $$L_1$$ is $$\mathbf{b}=(0,0,1)$$ and its equation is $$x=1,\;y=2,\;z=t$$.

Line $$L_2$$ passes through $$P_2(\lambda,5,6)$$ and is parallel to the $$y$$-axis.
Its direction vector is $$\mathbf{d}=(0,1,0)$$ and its equation is $$x=\lambda,\;y=5+s,\;z=6$$.

The shortest (perpendicular) distance $$D$$ between two skew lines $$\mathbf{r}=\mathbf{a}+t\mathbf{b}$$ and $$\mathbf{r}=\mathbf{c}+s\mathbf{d}$$ is given by
$$D=\frac{\lvert(\mathbf{a}-\mathbf{c})\cdot(\mathbf{b}\times\mathbf{d})\rvert}{\lvert\mathbf{b}\times\mathbf{d}\rvert}$$.

First find $$\mathbf{b}\times\mathbf{d}$$:
$$\mathbf{b}\times\mathbf{d}=\begin{vmatrix} \mathbf{i}&\mathbf{j}&\mathbf{k}\\ 0&0&1\\ 0&1&0 \end{vmatrix}=(-1,0,0).$$
Its magnitude is $$\lvert\mathbf{b}\times\mathbf{d}\rvert=1$$.

Next, $$\mathbf{a}-\mathbf{c}=P_1P_2=(1-\lambda,\,2-5,\;3-6)=(1-\lambda,-3,-3)$$.

Dot product:
$$(\mathbf{a}-\mathbf{c})\cdot(\mathbf{b}\times\mathbf{d})=(1-\lambda,-3,-3)\cdot(-1,0,0)=\lambda-1.$$

Therefore the shortest distance is
$$D=\lvert\lambda-1\rvert.$$

The question states this distance equals $$3$$, so
$$\lvert\lambda-1\rvert=3 \;\Longrightarrow\; \lambda=4 \text{ or } \lambda=-2.$$
Thus $$\lambda_1=4,\;\lambda_2=-2 \;(\lambda_2\lt\lambda_1).$$

The required point is $$(\lambda_1,\lambda_2,7)=(4,-2,7).$$ Let this point be $$Q(4,-2,7).$$

Distance of a point $$Q$$ from a line through $$P_1$$ with direction $$\mathbf{b}$$ is
$$\text{dist}=\frac{\lvert\overrightarrow{P_1Q}\times\mathbf{b}\rvert}{\lvert\mathbf{b}\rvert}.$$

Compute $$\overrightarrow{P_1Q}=(4-1,\,-2-2,\,7-3)=(3,-4,4).$$

Cross product:
$$\overrightarrow{P_1Q}\times\mathbf{b}=\begin{vmatrix} \mathbf{i}&\mathbf{j}&\mathbf{k}\\ 3&-4&4\\ 0&0&1 \end{vmatrix}=(-4,-3,0).$$

Magnitude:
$$\lvert(-4,-3,0)\rvert=\sqrt{(-4)^2+(-3)^2}= \sqrt{16+9}=5.$$

Because $$\lvert\mathbf{b}\rvert=1$$, the perpendicular distance is $$5$$.

Hence the square of this distance is $$5^2=25$$.

Therefore the required value is $$25$$, i.e. Option C.