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(C) The given lines are $L_1: \frac{x-\lambda}{2}=\frac{y-4}{3}=\frac{z-3}{4}$ and $L_2: \frac{x-2}{4}=\frac{y-4}{6}=\frac{z-7}{8}$.Note that the dire

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(C) The given lines are $L_1: \frac{x-\lambda}{2}=\frac{y-4}{3}=\frac{z-3}{4}$ and $L_2: \frac{x-2}{4}=\frac{y-4}{6}=\frac{z-7}{8}$.Note that the direction vectors are $\vec{b}_1 = 2\hat{i} + 3\hat{j} + 4\hat{k}$ and $\vec{b}_2 = 4\hat{i} + 6\hat{j} + 8\hat{k} = 2(2\hat{i} + 3\hat{j} + 4\hat{k})$.Since $\vec{b}_2 = 2\vec{b}_1$,the lines are parallel.The shortest distance between two parallel lines $\vec{r} = \vec{a}_1 + t\vec{b}$ and $\vec{r} = \vec{a}_2 + s\vec{b}$ is given by $d = \frac{|(\vec{a}_2 - \vec{a}_1) 	imes \vec{b}|}{|\vec{b}|}$.Here,$\vec{a}_1 = \lambda\hat{i} + 4\hat{j} + 3\hat{k}$,$\vec{a}_2 = 2\hat{i} + 4\hat{j} + 7\hat{k}$,and $\vec{b} = 2\hat{i} + 3\hat{j} + 4\hat{k}$.$\vec{a}_2 - \vec{a}_1 = (2-\lambda)\hat{i} + 0\hat{j} + 4\hat{k}$.$(\vec{a}_2 - \vec{a}_1) 	imes \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 2-\lambda & 0 & 4 \ 2 & 3 & 4 \end{vmatrix} = \hat{i}(0-12) - \hat{j}(8-12+6\lambda) + \hat{k}(6\lambda-6) = -12\hat{i} + (4-6\lambda)\hat{j} + (6\lambda-6)\hat{k}$.Magnitude $|(\vec{a}_2 - \vec{a}_1) 	imes \vec{b}| = \sqrt{(-12)^2 + (4-6\lambda)^2 + (6\lambda-6)^2} = \sqrt{144 + 16 - 48\lambda + 36\lambda^2 + 36\lambda^2 - 72\lambda + 36} = \sqrt{72\lambda^2 - 120\lambda + 196}$.$|\vec{b}| = \sqrt{2^2 + 3^2 + 4^2} = \sqrt{4+9+16} = \sqrt{29}$.Given $d = \frac{13}{\sqrt{29}}$,so $\frac{\sqrt{72\lambda^2 - 120\lambda + 196}}{\sqrt{29}} = \frac{13}{\sqrt{29}}$.$72\lambda^2 - 120\lambda + 196 = 169 \implies 72\lambda^2 - 120\lambda + 27 = 0$.Dividing by $3$,$24\lambda^2 - 40\lambda + 9 = 0$.Using the quadratic formula $\lambda = \frac{40 \pm \sqrt{1600 - 864}}{48} = \frac{40 \pm \sqrt{736}}{48}$.Wait,checking the calculation: $|12\hat{i} - 4\lambda\hat{j} + (3\lambda-6)\hat{k}| = 13 \implies 144 + 16\lambda^2 + 9\lambda^2 - 36\lambda + 36 = 169 \implies 25\lambda^2 - 36\lambda + 11 = 0$.$(25\lambda - 11)(\lambda - 1) = 0$. Thus $\lambda = 1$ or $\lambda = 11/25$.

If the shortest distance between the lines $\frac{x-\lambda}{2}=\frac{y-4}{3}=\frac{z-3}{4}$ and $\frac{x-2}{4}=\frac{y-4}{6}=\frac{z-7}{8}$ is $\frac{13}{\sqrt{29}}$,then a value of $\lambda$ is :

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