If the lines overrightarrow{r} = (hat{i} - ha | vedclass

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(B) The lines are given by $L_1: \overrightarrow{r} = (\hat{i} - \hat{j} + \hat{k}) + \lambda(3\hat{j} - \hat{k})$ and $L_2: \overrightarrow{r} = (\al

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$\frac{2}{11}$

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(B) The lines are given by $L_1: \overrightarrow{r} = (\hat{i} - \hat{j} + \hat{k}) + \lambda(3\hat{j} - \hat{k})$ and $L_2: \overrightarrow{r} = (\alpha\hat{i} - \hat{j}) + \mu(2\hat{i} - 3\hat{k})$.Since the lines are coplanar,the scalar triple product of the vector connecting the points on the lines and the direction vectors must be zero:$\begin{vmatrix} \alpha - 1 & 0 & -1 \ 0 & 3 & -1 \ 2 & 0 & -3 \end{vmatrix} = 0$Expanding the determinant: $(\alpha - 1)(-9) - 0 + (-1)(0 - 6) = 0$$-9\alpha + 9 + 6 = 0 \Rightarrow 9\alpha = 15 \Rightarrow \alpha = \frac{5}{3}$.The normal vector to the plane is $\overrightarrow{n} = (3\hat{j} - \hat{k}) 	imes (2\hat{i} - 3\hat{k}) = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 0 & 3 & -1 \ 2 & 0 & -3 \end{vmatrix} = -9\hat{i} + 2\hat{j} - 6\hat{k}$.The equation of the plane passing through $(1, -1, 1)$ is $-9(x - 1) + 2(y + 1) - 6(z - 1) = 0$,which simplifies to $9x - 2y + 6z - 17 = 0$.The distance from $(\frac{5}{3}, 0, 0)$ to the plane $9x - 2y + 6z - 17 = 0$ is $d = \frac{|9(\frac{5}{3}) - 2(0) + 6(0) - 17|}{\sqrt{9^2 + (-2)^2 + 6^2}} = \frac{|15 - 17|}{\sqrt{81 + 4 + 36}} = \frac{2}{\sqrt{121}} = \frac{2}{11}$.

If the lines $\overrightarrow{r} = (\hat{i} - \hat{j} + \hat{k}) + \lambda(3\hat{j} - \hat{k})$ and $\overrightarrow{r} = (\alpha\hat{i} - \hat{j}) + \mu(2\hat{i} - 3\hat{k})$ are coplanar,then the distance of the plane containing these two lines from the point $(\alpha, 0, 0)$ is

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