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The vector equation of the plane containing the lines $r=(\hat{\mathbf{i}}+\hat{\mathbf{j}})+t(\hat{\mathbf{i}}+2 \hat{\mathbf{j}}-\hat{\mathbf{k}})$ and $r=(\hat{\mathbf{i}}+\hat{\mathbf{j}})+s(-\hat{\mathbf{i}}+\hat{\mathbf{j}}-2 \hat{\mathbf{k}})$ is
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$\mathrm{r} \cdot \mathrm{n}=0$, where $\mathrm{n}=\hat{\mathrm{i}}-\hat{\mathrm{j}}-\hat{\mathrm{k}}$
The two given lines pass through the point
having position vector $\mathbf{a}=\hat{\mathbf{i}}+\hat{\mathbf{j}}$ and are parallel to the vectors $\mathbf{b}_1=\hat{\mathbf{i}}+2 \hat{\mathbf{j}}-\hat{\mathbf{k}}$ and $\mathbf{b}_2=-\hat{\mathbf{i}}+\hat{\mathbf{j}}-2 \hat{\mathbf{k}}$ respectively. Therefore, the plane containing the given lines also passes through the point with position vector $\mathbf{a}=\hat{i}+\hat{j}$. Since, the plane contains the lines which are parallel to the vectors $\mathbf{b}_1$ and $\mathbf{b}_2$ respectivley. Therefore, the plane is normal to the vector $\mathbf{n}$ given by.
$\begin{aligned} \mathbf{n} & =\mathbf{b}_1 \times \mathbf{b}_2=\left|\begin{array}{ccc}\hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ 1 & 2 & -1 \\ -1 & 1 & -2\end{array}\right| \\ & =\hat{\mathbf{i}}(-4+1)-\hat{\mathbf{j}}(-2-1)+\hat{\mathbf{k}}(1+2) \\ & =-3 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}+3 \hat{\mathbf{k}}\end{aligned}$
Thus, the vector equation of the required plane is $\mathbf{r} \cdot \mathbf{n}=\mathbf{a} \cdot \mathbf{n}$
$\begin{array}{lc}\Rightarrow & \mathbf{r} \cdot(-3 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}+3 \hat{\mathbf{k}})=(\hat{\mathbf{i}}+\hat{\mathbf{j}}) \cdot(-3 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}+3 \hat{\mathbf{k}}) \\ \Rightarrow & \mathbf{r} \cdot(-\hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}})=0 \Rightarrow \mathbf{r}(\hat{\mathbf{i}}-\hat{\mathbf{j}}-\hat{\mathbf{k}})=0\end{array}$
having position vector $\mathbf{a}=\hat{\mathbf{i}}+\hat{\mathbf{j}}$ and are parallel to the vectors $\mathbf{b}_1=\hat{\mathbf{i}}+2 \hat{\mathbf{j}}-\hat{\mathbf{k}}$ and $\mathbf{b}_2=-\hat{\mathbf{i}}+\hat{\mathbf{j}}-2 \hat{\mathbf{k}}$ respectively. Therefore, the plane containing the given lines also passes through the point with position vector $\mathbf{a}=\hat{i}+\hat{j}$. Since, the plane contains the lines which are parallel to the vectors $\mathbf{b}_1$ and $\mathbf{b}_2$ respectivley. Therefore, the plane is normal to the vector $\mathbf{n}$ given by.
$\begin{aligned} \mathbf{n} & =\mathbf{b}_1 \times \mathbf{b}_2=\left|\begin{array}{ccc}\hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ 1 & 2 & -1 \\ -1 & 1 & -2\end{array}\right| \\ & =\hat{\mathbf{i}}(-4+1)-\hat{\mathbf{j}}(-2-1)+\hat{\mathbf{k}}(1+2) \\ & =-3 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}+3 \hat{\mathbf{k}}\end{aligned}$
Thus, the vector equation of the required plane is $\mathbf{r} \cdot \mathbf{n}=\mathbf{a} \cdot \mathbf{n}$
$\begin{array}{lc}\Rightarrow & \mathbf{r} \cdot(-3 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}+3 \hat{\mathbf{k}})=(\hat{\mathbf{i}}+\hat{\mathbf{j}}) \cdot(-3 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}+3 \hat{\mathbf{k}}) \\ \Rightarrow & \mathbf{r} \cdot(-\hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}})=0 \Rightarrow \mathbf{r}(\hat{\mathbf{i}}-\hat{\mathbf{j}}-\hat{\mathbf{k}})=0\end{array}$
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