If vec{a} = hat{i} - 2hat{j} + 3hat{k},vec{b} | vedclass

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(C) Given vectors are $\vec{a} = \hat{i} - 2\hat{j} + 3\hat{k}$,$\vec{b} = 2\hat{i} + 3\hat{j} - \hat{k}$,and $\vec{c} = r\hat{i} + \hat{j} + (2r - 1)

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(C) Given vectors are $\vec{a} = \hat{i} - 2\hat{j} + 3\hat{k}$,$\vec{b} = 2\hat{i} + 3\hat{j} - \hat{k}$,and $\vec{c} = r\hat{i} + \hat{j} + (2r - 1)\hat{k}$.Since $\vec{c}$ is parallel to the plane of $\vec{a}$ and $\vec{b}$,the vectors $\vec{a}, \vec{b}$,and $\vec{c}$ are coplanar.For three vectors to be coplanar,their scalar triple product must be zero,i.e.,$[\vec{a} \, \vec{b} \, \vec{c}] = 0$.This is equivalent to the determinant of their components being zero:$\begin{vmatrix} 1 & -2 & 3 \ 2 & 3 & -1 \ r & 1 & 2r - 1 \end{vmatrix} = 0$Expanding the determinant along the first row:$1((3)(2r - 1) - (-1)(1)) - (-2)((2)(2r - 1) - (-1)(r)) + 3((2)(1) - (3)(r)) = 0$$1(6r - 3 + 1) + 2(4r - 2 + r) + 3(2 - 3r) = 0$$(6r - 2) + 2(5r - 2) + (6 - 9r) = 0$$6r - 2 + 10r - 4 + 6 - 9r = 0$$7r = 0$$r = 0$

If $\vec{a} = \hat{i} - 2\hat{j} + 3\hat{k}$,$\vec{b} = 2\hat{i} + 3\hat{j} - \hat{k}$ and $\vec{c} = r\hat{i} + \hat{j} + (2r - 1)\hat{k}$ are three vectors such that $\vec{c}$ is parallel to the plane of $\vec{a}$ and $\vec{b}$,then $r$ is equal to

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