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(C) The equation of line $L_1$ passing through $(1, 2, 3)$ and parallel to the $z$-axis is $\frac{x-1}{0} = \frac{y-2}{0} = \frac{z-3}{1}$.The equatio

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$25$

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(C) The equation of line $L_1$ passing through $(1, 2, 3)$ and parallel to the $z$-axis is $\frac{x-1}{0} = \frac{y-2}{0} = \frac{z-3}{1}$.The equation of line $L_2$ passing through $(\lambda, 5, 6)$ and parallel to the $y$-axis is $\frac{x-\lambda}{0} = \frac{y-5}{1} = \frac{z-6}{0}$.The shortest distance $SD$ between two lines $\vec{r} = \vec{a_1} + t\vec{b_1}$ and $\vec{r} = \vec{a_2} + s\vec{b_2}$ is given by $\frac{|(\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} 	imes \vec{b_2})|}{|\vec{b_1} 	imes \vec{b_2}|}$.Here,$\vec{a_1} = (1, 2, 3)$,$\vec{b_1} = (0, 0, 1)$,$\vec{a_2} = (\lambda, 5, 6)$,$\vec{b_2} = (0, 1, 0)$.$\vec{a_2} - \vec{a_1} = (\lambda-1, 3, 3)$.$\vec{b_1} 	imes \vec{b_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 0 & 0 & 1 \ 0 & 1 & 0 \end{vmatrix} = -\hat{i}$.$SD = \frac{|(\lambda-1, 3, 3) \cdot (-1, 0, 0)|}{|-1|} = |-(\lambda-1)| = |\lambda-1| = 3$.Thus,$\lambda-1 = 3$ or $\lambda-1 = -3$,so $\lambda = 4$ or $\lambda = -2$.Given $\lambda_2 The point is $P(4, -2, 7)$. Line $L_1$ is $(1, 2, z)$.The distance squared from $P(4, -2, 7)$ to $L_1$ is the distance to the point $(1, 2, 7)$ on $L_1$ (since the $z$-coordinate of $P$ is $7$,the projection onto $L_1$ is $(1, 2, 7)$).$PQ^2 = (4-1)^2 + (-2-2)^2 + (7-7)^2 = 3^2 + (-4)^2 + 0^2 = 9 + 16 = 25$.

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