Dimension of intersection of three subspaces

Question

Question: let $W_1, W_2, W_3$ be three distinct subspaces of vector space $\mathbb{R^{12}}$ each having dimension $4$, and let $W=W_1 ∩W_2 ∩ W_3$ then which of the following is/ are correct?

(1) $W$ is subspace of $\mathbb{R^{12}}$

(2) $dim(W)< 8$

(3) $dim(W)>7$

(4) $dim(w)<10$

My attempt: clearly $W$ will be subspace of $V$ (as intersection of subspaces is again subspace) hence (1) is true.

I stuck on other options, but I know $dim(W)≤dimV=12$. Further, I know if $W_1, W_2$ are subspaces of finite dimensional vector space then,

$dim(W_1+W_2)= dim(W_1)+dim(W_2)-dim(W_1∩W_2)$

But, how to use this formula, as we had intersection of three subspaces? further, using above formula is useful here? Please help needed.

Answer 1

$W$ is also a subspace of $W_i$, so $\dim W \leq \dim W_i = 4$.

Answer 2

$\dim W \leq \dim W_i = 4$ but since each $W_i$ are distinct $\dim W≠4$, therefore dimension of $W$ is strictly less than 4.Also notice that each subspace $W_1,W_2$ and $W_3$ can be chosen such that there intersection is $\{0\}$.Hence $$ 0≤\dim W \leq 3$$ Now obviously $W$ is subspace of $\mathbb{R}^{12}$, therefore correct options are $(1),(2)$ and $(4)$