If three subspaces in $R^3$ $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\} $ then find $dim(w_1 \cap w_2 \cap w_3), dim(w_1+w_2)$
My attempt : basis of 3 subspaces $B_{w_1} = (1,0,1)(0,1,1)\\B_{w_2} = (2,0,3)(0,2,1)\\ B_{w_3} = (3,0,-1)(0,3,7)\\$
Now $dim(w_1 + w_2) $ is obtained by row reduction of $w_1 \cup w_2$ and i got this reduced to (1,0,0) (0,1,0)(0,0,1). So $dim (w_1 + w_2) = 3$
This implies $dim(w_1 \cap w_2) = 2+2-3 = 1$ ---> (A)
But if i take in following way , i am getting different answer for $dim(w_1 \cap w_2)$
Let any arbitrary vector in $w_1$ is (a,b,a+b). If this is present in $w_2$ then
$(a,b,a+b ) = x(2,0,3)+y(0,2,1) \\ = (2x, 2y,3x+y) \\ \implies inconsistency$ for some scalars x,y
So $w_1 \cap w_2 = \phi \implies dim (w_1 \cap w_2) = 0$ But by set theory, i am getting it as 1. (from (A))
I'd just use some shortcuts.
For example, $\dim(W_1 + W_2) = 3$ because clearly, $\dim(W_1) = 2$ and there is an element in $W_2$ that is not in $W_1$, which one may adjoint to a basis of $W_1$ to obtain a basis of $\mathbb R^3$.
Your first computation of $\dim(W_1 \cap W_2)$ is based on the morphism $\mathbb R^3 \to (\mathbb R^3 / W_1) \times (\mathbb R^3 / W_2)$, whose kernel is indeed $W_1 \cap W_2$. The image is two-dimensional, and thus the kernel must be one-dimensional. Your solution is hence correct.
I don't see though how you derive your "inconsistency". For example, take $x = y$ and $a = b = x/2$.