The Cartesian equation of the plane vec{r}=(2 | vedclass

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(C) The given equation of the plane is in the form $\vec{r} = \vec{a} + \lambda \vec{b} + \mu \vec{c}$,where $\vec{a} = 2 \hat{i} - 3 \hat{j}$,$\vec{b

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$5 x-3 y-z=19$

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(C) The given equation of the plane is in the form $\vec{r} = \vec{a} + \lambda \vec{b} + \mu \vec{c}$,where $\vec{a} = 2 \hat{i} - 3 \hat{j}$,$\vec{b} = \hat{i} + 2 \hat{j} - \hat{k}$,and $\vec{c} = 2 \hat{i} + 3 \hat{j} + \hat{k}$.To find the Cartesian equation,we need the normal vector $\vec{n} = \vec{b} 	imes \vec{c}$.$\vec{n} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 1 & 2 & -1 \ 2 & 3 & 1 \end{vmatrix} = \hat{i}(2+3) - \hat{j}(1+2) + \hat{k}(3-4) = 5 \hat{i} - 3 \hat{j} - \hat{k}$.The Cartesian equation of the plane is given by $(\vec{r} - \vec{a}) \cdot \vec{n} = 0$,which is $(x-2, y+3, z-0) \cdot (5, -3, -1) = 0$.$5(x-2) - 3(y+3) - 1(z) = 0$.$5x - 10 - 3y - 9 - z = 0$.$5x - 3y - z = 19$.

The Cartesian equation of the plane $\vec{r}=(2 \hat{i}-3 \hat{j})+\lambda(\hat{i}+2 \hat{j}-\hat{k})+\mu(2 \hat{i}+3 \hat{j}+\hat{k})$ is

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