Is this set a basis of the vector space $H$? When $H=\{(x_1,x_2,x_3)^T\in\Bbb{R^3}:3x_1-2x_2+6x_3=0\}.$ I claim that the set $V=\{v_1 = (2,0,3), v_2 = (2,3,0), v_3 = (0,3,1)\}$ is a basis. My logic is this. This set is linearly independent, and it spans the entirety of $\Bbb{R^3}$. Thus it must also span each subspace, including $H$. Is this enough to $V$ is a basis? Are there other methods we could use?
 A: No, it is not enough. For a set to be a basis of, let's say, $V$, it has to be a linearly independent set whose span is $V$. For example, in $\mathbb{R}^2$, a line through the origin with a slope of 1: a basis would be $\{(1,1)\}$, not $\{(1,0), (0,1)\}$, since the span of the second set is all $\mathbb{R}^2$, not just the line itself.
A: The most immediate issue is that $v_1 \notin V$, which violates the fact that bases must be subsets of their spaces. This is important for all sorts of reasons, for example, we cannot say $\operatorname{span} \{v_1, v_2, v_3\} = V$, as $v_1$ lies in the left hand side, but not the right.
More intuitively, this should hopefully not make sense to you. If we have a basis of three vectors, then all bases for $V$ have three vectors, and $V$ should be a $3$-dimensional space. The definition of $V$ makes it clear that it's a plane, specifically the plane $3x - 2y + 6z = 0$ in general form. Planes are two-dimensional objects, so a basis ought to only contain $2$ vectors.
Instead, observe that $\{v_2, v_3\}$ is a linearly independent subset of $V$, hence $\dim V \ge 2$. Since $V$ is a proper subspace of $\Bbb{R}^3$, which has dimension $3$, then $\dim V \le 2$. Thus, $\dim V = 2$, and hence any linearly independent $2$-element subset of $V$ must be a basis.
A: No. The reason being the first vector of your claimed basis i.e. $v_1$ doesn't belong to the space $H$.

$3(2)-2(0)+6(3)=24\ne 0$

