Span of a Vector Space in $\mathbb{R}^3$ Consider the subspaces $W_1$ and $W_2$ of $\mathbb{R}^3$ given by
$W_1= \{(x,y,z) \in \mathbb{R}^3:x+y+z=0 \}$ and $W_2=\{(x,y,z) \in \mathbb{R}^3:x-y+z=0 \}$.
If $W$ is a subspace of $\mathbb{R}^3$ such that 


*

*$W \cap W_2= \mathrm{span}\bigl\{(0,1,1)\bigr\}$

*$W \cap W_1$ is orthogonal to $W \cap W_2$ with respect to the usual inner product of $\mathbb{R}^3$ 
then which of these are true?


*

*$W = \mathrm{span} \bigl\{ (0,1,-1),(0,1,1) \bigr\}$

*$W = \mathrm{span} \bigl\{ (1,0,-1),(0,1,-1) \bigr\}$

*$W = \mathrm{span} \bigl\{ (1,0,-1),(0,1,1) \bigr\}$

*$W = \mathrm{span} \bigl\{ (1,0,-1),(1,0,1) \bigr\}$
My Attempt:
$x+y+z=0 \implies x+y=-z$ so that free variables are two so $\mathrm{dim}(W_1)=2$ and similarly $x-y+z=0 \implies x+z=y$ so that $\mathrm{dim}(W_2)=2$.
Also $W \cap W_2 = \mathrm{span}\bigl\{(0,1,1)
\bigr\}$ implies $(0,1,1)$ is one element of $W$ so options 2,4 discarded. 
How to approach this type of problems in general?

You reasoned correctly and discarded $2$ and $4$.  It must be $1$, since $(1,0,-1)$ isn't orthogonal to $(0,1,1)$.
