We know that $\complex{3}$ is a vector space (Example VSCV). Consider the subset, \begin{equation*} W=\setparts{\colvector{x_1\\x_2\\x_3}}{2x_1-5x_2+7x_3=0} \end{equation*}
It is clear that $W\subseteq\complex{3}$, since the objects in $W$ are column vectors of size 3. But is $W$ a vector space? Does it satisfy the ten properties of Definition VS when we use the same operations? That is the main question.
Suppose $\vect{x}=\colvector{x_1\\x_2\\x_3}$ and $\vect{y}=\colvector{y_1\\y_2\\y_3}$ are vectors from $W$. Then we know that these vectors cannot be totally arbitrary, they must have gained membership in $W$ by virtue of meeting the membership test. For example, we know that $\vect{x}$ must satisfy $2x_1-5x_2+7x_3=0$ while $\vect{y}$ must satisfy $2y_1-5y_2+7y_3=0$. Our first property (Property AC) asks the question, is $\vect{x}+\vect{y}\in W$? When our set of vectors was $\complex{3}$, this was an easy question to answer. Now it is not so obvious. Notice first that \begin{equation*} \vect{x}+\vect{y}=\colvector{x_1\\x_2\\x_3}+\colvector{y_1\\y_2\\y_3}= \colvector{x_1+y_1\\x_2+y_2\\x_3+y_3} \end{equation*} and we can test this vector for membership in $W$ as follows. Because $\vect{x}\in W$ we know $2x_1-5x_2+7x_3=0$ and because $\vect{y}\in W$ we know $2y_1-5y_2+7y_3=0$. Therefore, \begin{align*} 2(x_1+y_1)-5(x_2+y_2)+7(x_3+y_3) &=2x_1+2y_1-5x_2-5y_2+7x_3+7y_3\\ &=(2x_1-5x_2+7x_3)+(2y_1-5y_2+7y_3)\\ &=0 + 0\\ &=0 \end{align*} and by this computation we see that $\vect{x}+\vect{y}\in W$. One property down, nine to go.
If $\alpha$ is a scalar and $\vect{x}\in W$, is it always true that $\alpha\vect{x}\in W$? This is what we need to establish Property SC. Again, the answer is not as obvious as it was when our set of vectors was all of $\complex{3}$. Let us see. First, note that because $\vect{x}\in W$ we know $2x_1-5x_2+7x_3=0$. Therefore, \begin{equation*} \alpha\vect{x}=\alpha\colvector{x_1\\x_2\\x_3}=\colvector{\alpha x_1\\\alpha x_2\\\alpha x_3} \end{equation*} and we can test this vector for membership in $W$. First, note that because $\vect{x}\in W$ we know $2x_1-5x_2+7x_3=0$. Therefore, \begin{align*} 2(\alpha x_1)-5(\alpha x_2)+7(\alpha x_3) &=\alpha(2x_1-5x_2+7x_3)\\ &=\alpha 0\\ &=0 \end{align*} and we see that indeed $\alpha\vect{x}\in W$. Always.
If $W$ has a zero vector, it will be unique (Theorem ZVU). The zero vector for $\complex{3}$ should also perform the required duties when added to elements of $W$. So the likely candidate for a zero vector in $W$ is the same zero vector that we know $\complex{3}$ has. You can check that $\zerovector=\colvector{0\\0\\0}$ is a zero vector in $W$ too (Property Z).
With a zero vector, we can now ask about additive inverses (Property AI). As you might suspect, the natural candidate for an additive inverse in $W$ is the same as the additive inverse from $\complex{3}$. However, we must insure that these additive inverses actually are elements of $W$. Given $\vect{x}\in W$, is $\vect{-x}\in W$? \begin{equation*} \vect{-x}=\colvector{-x_1\\-x_2\\-x_3} \end{equation*} and we can test this vector for membership in $W$. As before, because $\vect{x}\in W$ we know $2x_1-5x_2+7x_3=0$. \begin{align*} 2(-x_1)-5(-x_2)+7(-x_3) &=-(2x_1-5x_2+7x_3)\\ &=-0\\ &=0 \end{align*} and we now believe that $\vect{-x}\in W$.
Is the vector addition in $W$ commutative (Property C)? Is $\vect{x}+\vect{y}=\vect{y}+\vect{x}$? Of course! Nothing about restricting the scope of our set of vectors will prevent the operation from still being commutative. Indeed, the remaining five properties are unaffected by the transition to a smaller set of vectors, and so remain true. That was convenient.
So $W$ satisfies all ten properties, is therefore a vector space, and thus earns the title of being a subspace of $\complex{3}$.