Answer
The subspace contains only the zero vector and hence there is no Basis
There are no non-zero vectors present, therefore the dimension for the given subspace is 0
Work Step by Step
We are required to find the basis and state the dimension:
Let the given subspace be represented by $\mathbf{{S}}$ such that:
$\mathbf{S}=\begin{pmatrix}a-3b+c=0\\b-2c=0\\2b-c=0\end{pmatrix} :(a,b,c)$
Writing the given set in matrix form;
$\mathbf{S}=\begin{bmatrix}1&-3&1\\0&1&-2\\0&2&-1\end{bmatrix} \begin{bmatrix}a\\b\\c\end{bmatrix} =\begin{bmatrix}0\\0\\0\end{bmatrix}$
We form an augmented matrix $\mathbf{M}$ and row reduce$\mathbf{ [A\,0] }$to echelon form ,
$\begin{bmatrix}1&-3&1&0\\0&1&-2&0\\0&2&-1&0\end{bmatrix} \sim\begin{bmatrix}1&0&0&0\\0&1&0&0\\0&0&1&0\end{bmatrix}$
This shows that the subspace contains only the zero vector and hence there is no Basis
There are no non-zero vectors present, therefore the dimension for the given subspace is 0
