If the vector equation of the plane bar{r}=(2 | vedclass

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(A) The given equation of the plane is in the form $\bar{r} = \bar{a} + \lambda \bar{b} + \mu \bar{c}$,where $\bar{a} = 2 \hat{i} + \hat{k}$,$\bar{b}

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$2$

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(A) The given equation of the plane is in the form $\bar{r} = \bar{a} + \lambda \bar{b} + \mu \bar{c}$,where $\bar{a} = 2 \hat{i} + \hat{k}$,$\bar{b} = \hat{i}$,and $\bar{c} = \hat{i} + 2 \hat{j} - 3 \hat{k}$.The normal vector $\bar{n}$ to the plane is given by the cross product $\bar{b} 	imes \bar{c}$:$\bar{n} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 1 & 0 & 0 \ 1 & 2 & -3 \end{vmatrix} = \hat{i}(0 - 0) - \hat{j}(-3 - 0) + \hat{k}(2 - 0) = 3 \hat{j} + 2 \hat{k}$.The scalar product form of the plane equation is $\bar{r} \cdot \bar{n} = \bar{a} \cdot \bar{n}$.Calculating $\bar{a} \cdot \bar{n}$:$\bar{a} \cdot \bar{n} = (2 \hat{i} + \hat{k}) \cdot (3 \hat{j} + 2 \hat{k}) = (2)(0) + (0)(3) + (1)(2) = 2$.Comparing this with $\bar{r} \cdot (3 \hat{j} + 2 \hat{k}) = \alpha$,we get $\alpha = 2$.

If the vector equation of the plane $\bar{r}=(2 \hat{i}+\hat{k})+\lambda \hat{i}+\mu(\hat{i}+2 \hat{j}-3 \hat{k})$ in scalar product form is given by $\bar{r} \cdot(3 \hat{j}+2 \hat{k})=\alpha$,then $\alpha=$

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