<s>[INST] What is the dimension of $V$? Let $$A=\begin{pmatrix} 1 & 1& 1\\2 &2&3\\x&y&z\end{pmatrix}$$
and let $V=\{(x,y,z)\in \mathbb{R}^3: \det(A)=0\}$ Then the dimension of $V$ equals,
Efforts:
Applying row transformation $R_2\to R_2-2R_1$, we get $$A=\begin{pmatrix} 1 & 1& 1\\0 &0&1\\x&y&z\end{pmatrix}$$
Now expanding along the second row, I get $\det(A)=y-x$ so informally speaking there are two independent variables. 
Hence dimension is 2
Am I right?
Edits: maybe I have received downvote because of poor mathematical language. I meant $x$ is equal to $y$, so we are free to choose $x$ and $z$. Then we can write the basis as well.
 [/INST] You should be more precise with the language: “informally speaking” is not enough.
You can go on with the row reduction:
$$
\begin{pmatrix} 1 & 1& 1\\2 &2&3\\x&y&z\end{pmatrix}
\xrightarrow{\begin{aligned}R_2&\gets R_2-2R_1 \\ R_3&\gets R_3-xR_1\end{aligned}}
\begin{pmatrix} 1 & 1& 1\\0 &0&1\\0&y-x&z-x\end{pmatrix}
\xrightarrow{R_3\gets R_3-(z-x)R_2}
\begin{pmatrix} 1 & 1& 1\\0 &0&1\\0&y-x&0\end{pmatrix}
$$
None of these operation changes the determinant. Switching rows 2 and 3 and changing sign to the second row to compensate, we get a triangular form
\begin{pmatrix} 1 & 1& 1\\0&x-y&0\\0 &0&1\end{pmatrix}
and the determinant is $x-y$, so the condition is $x=y$. The variables $y$ and $z$ are “free”. You get two linearly independent solutions by first setting $y=1$ and $z=0$ and then $y=0$ and $z=1$.
Therefore a basis for the subspace you're given is
$$
\left\{
\begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix},
\begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix}
\right\}
$$
</s>