(1) A = { (a,b,2a-3b) | a,b $\in $ R }
$ \quad$ $A = span\{ (1,0,2) , (0,1,-3) \}\ {\rm is\ a\ subspace\ of}\ \mathbb{R}^3$
(2) $B = \{ (u,v,w) | vw = 0 $ }
**$ \quad$ $ Let B = span(T).\ (1,1,0) , (1,0,1) \in B.\ (1,1,0) + (1,0,1) = (2, 1, 1) \in span(T)$,
${\rm but\ this\ is\ impossible\ as\ (2,1,1) \not \in B\ since\ vw \ne 0.\ B\ is\ not\ a\ subspace.} $**
(3) C = { ($d^2,e^2, f^2)$ | d,e, f $\in $ R }
$ \quad$ $ Let \ d, e, f = 1, and\ C = span(T).\ (1,1,1) \in span(T) , then\ -1(1,1,1) = (-1,-1,-1) \in span(T),\ but\ this\ is\ impossible\ as\ -1 \ne d^2, e^2, f^2 since\ for\ any\ d, e, f, \in R , d^2, e^2, f^2 \ge 0\ but\ -1 < 0.\ \therefore C\ is\ not\ a\ subspace.\ $
(4) D = { (p,q,r) | $ \begin{pmatrix} p & q & r \\ 1 & 2 & 0 \\ 0 & 1 & 2\end{pmatrix}$ is singular }
$ \quad$ $ \begin{vmatrix} p & q & r \\ 1 & 2 & 0. \\ 0 & 1 & 2 \\ \end{vmatrix} $ = $ 4p - 2q + r = 0.$ $ \quad (p, q ,r) = (s-t, 2s, 4t) = s(1,2,0) + t(-1,0,4)\ where\ s,t \in \ R. $ $ \quad D = span\{ (1,2,0) , (-1,0,4) \} is\ a\ subspace\ of R^3 $
(5) E = { (x,y,z) | $ \begin{pmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9\end{pmatrix}$ $ \begin{pmatrix} x \\ y \\ z \end{pmatrix} $ = $ \begin{pmatrix} 2x \\ 2y \\ 2z \end{pmatrix} $ }
$ \quad$ $ x+2y+3z = 2x, 4x+5y+6z = 2y, 7x+8y + 9z = 2z. Gauss\ Jordan\ Elimination\ gives\ $
$ \quad unique\ soln : (x,y,z) = (0,0,0).\ \therefore \ E = span\{ (0,0,0) \} is\ a\ subspace\ of R^3 $
I am not sure about my reasons for (4) and (5), please correct me if I am wrong. Thank you.
For $4)$ you have the plane $4p-2q+r=0$ through the origin. That's a subspace.
For $5$, we have the kernel of $A-2I$, so it's a subspace.