Solution: Note that \begin{align*} W_1 & = \left\{ \left( x_1,x_2,x_3 \right)\in \mathbb{R} ^3 : x_1+2x_2-x_3 = 0 \right\} \\ & = \left\{ \left( x_1,x_2,x_3 \right) \in \mathbb{R} ^3 : x_1 - x_3 = 2x_2 \right\} \\ & = \left\{ \left( x_1,\frac{x_1 - x_3}{2}, x_3 \right) : x_1,x_3\in \mathbb{R} \right\} \\ & = \text{span}\left\{ \left( 2,1,0 \right), \left( 0,-1,2 \right) \right\} . \end{align*} Therefore, the dimension of $W_1$ is $2$. Similarly, note that \begin{align*} W_2 & = \left\{ \left( x_1,x_2,x_3 \right) \in \mathbb{R} ^3: 2x_1 =-3x_3 \right\} \\ & = \left\{ \left( x_1,x_2, -\frac{2}{3}x_1 \right) : x_1,x_2\in \mathbb{R} \right\} \\ & = \text{span} \left\{ \left( 3,0,-2 \right), \left( 0,1,0 \right) \right\}. \end{align*} Hence, the dimension of $W_2$ is also $2$.

We observe that the rank of the matrix \begin{align*} \begin{pmatrix} 2 & 1 & 0 \\ 0 & -1 & 2 \\ 3 & 0 & -2 \\ 0 & 1 & 0 \\ \end{pmatrix} \end{align*} is $3$ and hence the dimension of $W_1+W_2$ will be $3$. Since \begin{align*} & \dim \left( W_1 + W_2 \right) = \dim W_1 + \dim W_2 - \dim \left( W_1 \cap W_2 \right) \\ \implies & \dim \left( W_1 \cap W_2 \right) = 4-3 = 1 \end{align*}