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(A) The lines are given by $\vec{r_1} = (1, 2, 3) + t(2, 3, 4)$ and $\vec{r_2} = (\lambda, 4, 5) + s(3, 4, 5)$.The shortest distance $d$ between two l

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$\frac{5 \sqrt{2}}{3}$

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(A) The lines are given by $\vec{r_1} = (1, 2, 3) + t(2, 3, 4)$ and $\vec{r_2} = (\lambda, 4, 5) + s(3, 4, 5)$.The shortest distance $d$ 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 $d = \frac{|(\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} 	imes \vec{b_2})|}{|\vec{b_1} 	imes \vec{b_2}|}$.Here,$\vec{b_1} 	imes \vec{b_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 2 & 3 & 4 \ 3 & 4 & 5 \end{vmatrix} = \hat{i}(15-16) - \hat{j}(10-12) + \hat{k}(8-9) = -\hat{i} + 2\hat{j} - \hat{k}$.The magnitude is $|\vec{b_1} 	imes \vec{b_2}| = \sqrt{(-1)^2 + 2^2 + (-1)^2} = \sqrt{6}$.$\vec{a_2} - \vec{a_1} = (\lambda-1, 2, 2)$.Shortest distance $d = \frac{|(\lambda-1)(-1) + 2(2) + 2(-1)|}{\sqrt{6}} = \frac{|-\lambda + 1 + 4 - 2|}{\sqrt{6}} = \frac{|3-\lambda|}{\sqrt{6}}$.Given $d = \frac{1}{\sqrt{6}}$,so $|3-\lambda| = 1$,which gives $\lambda = 4$ or $\lambda = 2$.Thus,$\lambda_1 = 4$ and $\lambda_2 = 2$.We need the radius of the circle passing through $(0,0), (4,2)$ and $(2,4)$.Let the points be $O(0,0), A(4,2), B(2,4)$. The area of $	riangle OAB$ is $\Delta = \frac{1}{2} |0(2-4) + 4(4-0) + 2(0-2)| = \frac{1}{2} |16 - 4| = 6$.The side lengths are $OA = \sqrt{4^2+2^2} = \sqrt{20}$,$OB = \sqrt{2^2+4^2} = \sqrt{20}$,$AB = \sqrt{(4-2)^2 + (2-4)^2} = \sqrt{8}$.The radius $R = \frac{abc}{4\Delta} = \frac{\sqrt{20} \cdot \sqrt{20} \cdot \sqrt{8}}{4 \cdot 6} = \frac{20 \cdot 2\sqrt{2}}{24} = \frac{40\sqrt{2}}{24} = \frac{5\sqrt{2}}{3}$.

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