Solution: Note that \begin{align*} W_1 & = \left\{ \mathbf{x} \in \mathbb{R} ^3 : x_1 + 2x_2 - x_3 = 0 \right\} \\ & = \left\{ \mathbf{x} \in \mathbb{R} ^3: x_1 + 2x_2 = x_3 \right\} \\ & = \left\{ \left( x_1, x_2, x_1 + 2x_2 \right) : x_1, x_2 \in \mathbb{R} \right\} \\ & = \operatorname{span}\left\{ (1,0,1), (0,1,2) \right\} . \end{align*} Similarly, \begin{align*} W_2 & = \left\{ \mathbf{x} \in \mathbb{R} ^3 : 2x_1 + 3x_3 = 0 \right\}\\ & = \left\{ \left( x_1, x_2, -\frac{3}{2}x_1 \right) : x_1, x_2 \in \mathbb{R} \right\} \\ & = \operatorname{span}\left\{ (2,0,-3), (0,1,0) \right\} . \end{align*} Therefore, \[ \dim (W_1) = 2 = \dim (W_2). \]

Now, we also know that for any subspaces $W_1, W_2$ we must have \begin{align*} \dim \left( W_1 + W_2 \right) = \dim (W_1) + \dim (W_2) - \dim (W_1 \cap W_2). \end{align*} Note that \[ \det \begin{bmatrix} 1 & 0 & 1 \\ 0 & 1 & 2 \\ 0 & 1 & 0 \\ \end{bmatrix} = -2 \neq 0, \] thus, the set \[ \left\{ (1,0,1), (0,1,2), (0,1,0) \right\} \] is linearly independent and hence forms a basis for $\mathbb{R} ^3$. Since \[ W_1 + W_2 = \operatorname{span}\left\{ W_1 \cup W_2 \right\} \implies \dim (W_1 + W_2) = 3. \] Hence, \begin{align*} \dim (W_1 \cap W_2) = \dim (W_1) + \dim (W_2) - \dim \left( W_1 + W_2 \right) = 1. \end{align*} Thus, the correct options are (A), (B), (D) and (C) is FALSE.