i have de next problem
Let 
be a basis of R3 and 
. If v and w∈R3 satisfy 
, then ⟨v, w⟩∩S is equal to
and they give me the following options
I'll tell you what I did (I'm sure I did it wrong haha). try looking for (v + w) b = (α, β, ɣ) so v + w = 1 (1,0,0) +3 (-1,1,0) -1 (-1, -1,1) which gave me that v + w = (- 1,4, -1) I did the same with v-w and it gave me (-10, -2,5) then assume that v + w + v-w = (- 1,4, -1) + (- 10, -2,5) from there take v = (- 11 / 2,1,2) and w = (9 / 2,3, -3) but I'm very far from all the results (from the options they give me to choose from)
We know that $S=\langle \begin{pmatrix} 2\\1\\0 \end{pmatrix},\begin{pmatrix} 1\\0\\1 \end{pmatrix} \rangle$ and $\langle v,w\rangle=\langle \begin{pmatrix} -11\\2\\4 \end{pmatrix},\begin{pmatrix} 9\\6\\-6 \end{pmatrix} \rangle$, so, if we want to find the generator of $S \cap\langle v,w\rangle$ we need to solve the system $$\alpha\begin{pmatrix} 2\\1\\0 \end{pmatrix}+\beta\begin{pmatrix} 1\\0\\1 \end{pmatrix}=\gamma\begin{pmatrix} -11\\2\\4 \end{pmatrix}+\delta\begin{pmatrix} 9\\6\\-6 \end{pmatrix}\text{, with }\alpha,\beta,\gamma,\delta\in \mathbb R.$$ Starting from the matrix associated to the homogeneous system $\begin{pmatrix} 2&&1&&11&&-9\\1&&0&&-2&&-6\\0&&1&&-4&&6 \end{pmatrix}$
we get $\gamma=-\dfrac{20}{17}\delta$ $\implies S\cap\langle v,w \rangle$, whose dimension is $1$, is given by $$\langle -\dfrac{20}{17}\begin{pmatrix} -11\\2\\4 \end{pmatrix}+\begin{pmatrix} 9\\6\\-6 \end{pmatrix} \rangle=\langle\begin{pmatrix} -\dfrac{23}{17}\\-\dfrac{20}{17}\\1 \end{pmatrix}\rangle=\langle \begin{pmatrix} 23\\20\\-17 \end{pmatrix}\rangle.$$