Answer:
Given lines are \[\vec{r}=\hat{i}+\hat{j}+\lambda \,(\hat{i}+2\hat{j}-\hat{k})\] ? (i) \[\vec{r}=\hat{i}+\hat{j}+\mu \,\,(-\,\hat{i}+\hat{j}-2\hat{k})\] ? (ii) On comparing Eqs. (i) and (ii) with \[\vec{r}={{\vec{a}}_{1}}+\lambda {{\vec{b}}_{1}}\] and \[\vec{r}={{\vec{a}}_{2}}+\mu {{\vec{b}}_{2}},\] respectively we get \[{{\vec{a}}_{1}}=\hat{i}+\hat{j},\] \[{{\vec{b}}_{1}}=\,\hat{i}+2\hat{j}-\hat{k}\] and \[{{\vec{a}}_{2}}=\,\hat{i}+\hat{j},\] \[{{\vec{b}}_{2}}=-\,\hat{i}+\hat{j}-2\hat{k}\] \[\because \] Required plane contains lines (i) and (ii). \[\therefore \] The plane is parallel to the vectors \[{{\vec{b}}_{1}}\] and \[{{\vec{b}}_{2}}\]. So. Normal to the plane is parallel to the vector \[{{\vec{b}}_{1}}\times {{\vec{b}}_{2}}.\] Now, \[{{\vec{b}}_{1}}\times {{\vec{b}}_{2}}=\left| \begin{matrix} {\hat{i}} & {\hat{j}} & {\hat{k}} \\ 1 & 2 & -\,1 \\ -\,1 & 1 & -\,2 \\ \end{matrix} \right|\] \[=\hat{i}(-\,4+1)-\hat{j}(-\,2-\,1)+\hat{k}(1+2)\] \[=-\,3\hat{i}+3\hat{j}+3\hat{k}\] ? (iii) Thus, required plane passes through \[{{\vec{a}}_{1}}=\hat{i}+\hat{j}\] and having normal parallel to the vector \[{{\vec{b}}_{1}}\times {{\vec{b}}_{2}}.\] Now, its equation is given by \[(\vec{r}-{{\vec{a}}_{1}})\cdot ({{\vec{b}}_{1}}\times {{\vec{b}}_{2}})=0\] \[\Rightarrow \] \[[\vec{r}-(\hat{i}+\hat{j})]\cdot (-\,3\hat{i}+3\hat{j}\,+\hat{k})=0\] \[\Rightarrow \] \[\vec{r}[-\,3\hat{i}+3\hat{j}\,+3\hat{k}]+[-(\hat{i}+\hat{j})\] \[(-\,3\hat{i}\,+3\hat{j}\,+3\hat{k})]=0\] \[\Rightarrow \] \[\vec{r}\cdot [-\,3\hat{i}+3\hat{j}\,+3\hat{k}]+\] \[[(-\,1)(-\,3)+(-\,1)(3)+(0)(3)]=0\] \[\Rightarrow \] \[\vec{r}\cdot (-\,3\hat{i}+3\hat{j}+3\hat{k})+3-3=0\] \[\Rightarrow \] \[\vec{r}\cdot [3(-\,\hat{i}+\hat{j}+\hat{k})]=0\] \[\Rightarrow \] \[\vec{r}\cdot (-\,\hat{i}+\hat{j}+\hat{k})=0\] ?(iv) Its Cartesian form is \[-\,x+y+z=0\] ? (v) \[[put\,\,\vec{r}=x\hat{i}+y\hat{j}+z\hat{k}]=0\] This plane passes through the point (0, 0, 0). \[\therefore \] Distance of plane (v) from origin = 0 and distance of plane from the point (2, 2, 2) \[=\left| \frac{2\,(-\,1)+2\,(1)+2\,(1)}{\sqrt{{{(-\,1)}^{2}}+{{(1)}^{2}}+{{(1)}^{2}}}} \right|=\frac{2}{\sqrt{3}}\] units
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