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(C) The given lines are:$L_{1}: \frac{x-1}{2} = \frac{y-2}{1} = \frac{z-1}{3}$. Point $P_{1} = (1, 2, 1)$,direction vector $\vec{v}_{1} = 2\hat{i} + \

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$3$

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(C) The given lines are:$L_{1}: \frac{x-1}{2} = \frac{y-2}{1} = \frac{z-1}{3}$. Point $P_{1} = (1, 2, 1)$,direction vector $\vec{v}_{1} = 2\hat{i} + \hat{j} + 3\hat{k}$.$L_{2}: \frac{x-2}{1} = \frac{y-\lambda}{2} = \frac{z-3}{4}$. Point $P_{2} = (2, \lambda, 3)$,direction vector $\vec{v}_{2} = \hat{i} + 2\hat{j} + 4\hat{k}$.Let $\vec{a} = P_{2} - P_{1} = (2-1)\hat{i} + (\lambda-2)\hat{j} + (3-1)\hat{k} = \hat{i} + (\lambda-2)\hat{j} + 2\hat{k}$.The shortest distance $d$ is given by $d = \frac{|\vec{a} \cdot (\vec{v}_{1} 	imes \vec{v}_{2})|}{|\vec{v}_{1} 	imes \vec{v}_{2}|}$.First,calculate $\vec{v}_{1} 	imes \vec{v}_{2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 2 & 1 & 3 \ 1 & 2 & 4 \end{vmatrix} = \hat{i}(4-6) - \hat{j}(8-3) + \hat{k}(4-1) = -2\hat{i} - 5\hat{j} + 3\hat{k}$.Magnitude $|\vec{v}_{1} 	imes \vec{v}_{2}| = \sqrt{(-2)^2 + (-5)^2 + 3^2} = \sqrt{4 + 25 + 9} = \sqrt{38}$.Now,$\vec{a} \cdot (\vec{v}_{1} 	imes \vec{v}_{2}) = (1)(-2) + (\lambda-2)(-5) + (2)(3) = -2 - 5\lambda + 10 + 6 = 14 - 5\lambda$.Given $d = \frac{1}{\sqrt{38}}$,so $\frac{|14 - 5\lambda|}{\sqrt{38}} = \frac{1}{\sqrt{38}}$.$|14 - 5\lambda| = 1$.Case $1$: $14 - 5\lambda = 1 \Rightarrow 5\lambda = 13 \Rightarrow \lambda = 2.6$.Case $2$: $14 - 5\lambda = -1 \Rightarrow 5\lambda = 15 \Rightarrow \lambda = 3$.Since we need the integral value of $\lambda$,the answer is $3$.

If the shortest distance between the straight lines $3(x-1)=6(y-2)=2(z-1)$ and $4(x-2)=2(y-\lambda)=(z-3)$,$\lambda \in R$ is $\frac{1}{\sqrt{38}}$,then the integral value of $\lambda$ is equal to :

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