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

To say that $$\vec{c}$$ is parallel to the plane determined by $$\vec{a}$$ and $$\vec{b}$$ means that $$\vec{c}$$ is orthogonal to the normal of that plane.
The normal vector of the plane is given by the cross-product $$\vec{a}\times\vec{b}$$.

First find $$\vec{a}\times\vec{b}$$.
$$ \vec{a}\times\vec{b}= \begin{vmatrix} \hat{i} & \hat{j} & \hat{k}\\ 1 & -2 & 3\\ 2 & 3 & -1 \end{vmatrix} = \hat{i}\Big((-2)(-1)-3\cdot3\Big) -\hat{j}\Big(1\cdot(-1)-3\cdot2\Big) +\hat{k}\Big(1\cdot3-(-2)\cdot2\Big) $$ $$ = \hat{i}(2-9)\;-\;\hat{j}(-1-6)\;+\;\hat{k}(3+4) = -7\hat{i}+7\hat{j}+7\hat{k}. $$

Thus $$\vec{a}\times\vec{b}=(-7,\,7,\,7).$$

For orthogonality, the dot product of $$\vec{c}$$ with this normal must vanish: $$ \vec{c}\cdot(\vec{a}\times\vec{b})=0. $$ Compute the dot product: $$ (r,\,1,\,2r-1)\cdot(-7,\,7,\,7)= -7r + 7 + 7(2r-1). $$ $$ -7r + 7 + 14r - 7 = 7r. $$

Set this equal to zero: $$7r = 0 \;\Longrightarrow\; r = 0.$

Therefore, $$r = 0.$$

Option C which is: $$0$$