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If $$W_{1}=\{(x,y,z)| x+y-z=0\}$$; $$W_{2}=\{(x,y,z)| 3x+y-2z=0\}$$; $$W_{3}=\{(x,y,z)| x-7y+3z=0\}$$

Find the $\dim(W_{1}\cap W_{2}\cap W_{3}).$

My solution- I know how to find $\dim(W_{1}\cap W_{2})=\dim W_{1} + \dim W_{2}-\dim(W_{1}+W_{2}).$ But can we extend this further or is there any simpler way to find this.

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    To extend it you can define $V:=W_1\cap W_2$ and apply your result to $V\cap W_3$. Another way to do it is the one shwed in the answer – 2017-02-14
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    @Giulio Thanks! So, going by the following solution answer would be 2 itself ? I couldn't quite understand what's the need of Rank Nullity Theorem as answered by egreg. – 2017-02-14
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    The intersection will have dimension $n-r(A) = 3-2=1$ where $A$ is the matrix and $n$ is the number of variables. – 2017-02-14

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The intersection is the solution space of $$ \begin{cases} x+y-z=0\\ 3x+y-2z=0\\ x-7y+3z=0 \end{cases} $$ that is, the null space of the matrix $$ \begin{bmatrix} 1 & 1 & -1 \\ 3 & 1 & -2 \\ 1 & -7 & 3 \end{bmatrix} $$ which has rank $2$, as an easy elimination will prove. The rank-nullity theorem now helps.

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    Oh..yup thanks, a lot simpler way. But, the answer $dim(W_1 \cap W_2 \cap W_3)$ would be 2 itself. Whats the use of rank nullity theorem ? – 2017-02-14
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    @user18900 Since the rank is $2$, the null space has dimension $3-2=1$. – 2017-02-14