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(B) Since $(3,5,7)$ lies on $L_{1}$,we have $\frac{3-a}{l}=\frac{5-2}{3}=\frac{7-b}{4}=1$. From this,$3-a=l \Rightarrow a+l=3$ and $7-b=4 \Rightarrow

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$\frac{1}{\sqrt{6}}$

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(B) Since $(3,5,7)$ lies on $L_{1}$,we have $\frac{3-a}{l}=\frac{5-2}{3}=\frac{7-b}{4}=1$. From this,$3-a=l \Rightarrow a+l=3$ and $7-b=4 \Rightarrow b=3$. The vector from $(3,5,7)$ to $(4,3,8)$ is $\vec{v}_{1} = (4-3, 3-5, 8-7) = (1, -2, 1)$. The direction vector of $L_{1}$ is $\vec{v}_{2} = (l, 3, 4)$. Since the line segment is perpendicular to $L_{1}$,$\vec{v}_{1} \cdot \vec{v}_{2} = 0 \Rightarrow l - 6 + 4 = 0 \Rightarrow l = 2$. Thus,$a = 3 - 2 = 1$. The lines are $L_{1}: \frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}$ and $L_{2}: \frac{x-2}{3}=\frac{y-4}{4}=\frac{z-5}{5}$. Let $A = (1,2,3)$ and $B = (2,4,5)$,so $\vec{AB} = (1,2,2)$. The direction vectors are $\vec{p} = (2,3,4)$ and $\vec{q} = (3,4,5)$. $\vec{p} 	imes \vec{q} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 2 & 3 & 4 \ 3 & 4 & 5 \end{vmatrix} = \hat{i}(15-16) - \hat{j}(10-12) + \hat{k}(8-9) = -\hat{i} + 2\hat{j} - \hat{k}$. The shortest distance is $d = \left| \frac{\vec{AB} \cdot (\vec{p} 	imes \vec{q})}{|\vec{p} 	imes \vec{q}|} ight| = \left| \frac{(1,2,2) \cdot (-1,2,-1)}{\sqrt{(-1)^2 + 2^2 + (-1)^2}} ight| = \left| \frac{-1+4-2}{\sqrt{6}} ight| = \frac{1}{\sqrt{6}}$.

If the foot of the perpendicular from point $(4,3,8)$ on the line $L_{1}: \frac{x-a}{l}=\frac{y-2}{3}=\frac{z-b}{4},$ $l \neq 0$ is $(3,5,7),$ then the shortest distance between the line $L_{1}$ and line $L_{2}: \frac{x-2}{3}=\frac{y-4}{4}=\frac{z-5}{5}$ is equal to:

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