Most introductory books on Linear Algebra have a Theorem which says something like
Let $A$ be a square $n \times n$ matrix. Then the following are equivalent:
What does this mean, it simply means that if you want to check if any of these conditions is true or false, you can simply pick whichever other condition from the list and check it instead..
Your question is: Can instead of third or fourth condition, check the second? That's exactly what the Theorem says: YES.
If $\big\{\vec v_1= (1,1,1), \, \vec v_2= (0,0,1),\, \vec v_3=(0,1,1)\big\}$ form a basis of $\mathbb R^3$, then you can show that any $(x,y,z)\in \mathbb R^3$ can be written as $(x,y,z)=x\cdot \vec v_1 +(z-y) \cdot\vec v_2 +(y-x) \cdot \vec v_3$.
So, $T(x,y,z)\begin{array}[t]{l}= x\cdot T(\vec v_1) + (z-y)\cdot T(\vec v_2) + (y-x)\cdot T(\vec v_3)\\=
x\cdot(1,2,0)+(z-y)\cdot (2,8,8) +(y-x)\cdot (1,5,6)
\\
=(-y+2z,\,-3x-3y+8z,\, -6x-2y+8z)\end{array}$
$\ker T=\big\{(x,y,z)\in \mathbb R^3:T(x,y,z)=(0,0,0)\big\}=\ldots= \big\{\left(x,3x,\frac {3x}{2}\right):x \in \mathbb R\big\}\\\implies \dim \ker T=1$
$ImT=\big\{(X,Y,Z)\in \mathbb R^3:X=-y+2z,\, Y=-3x-3y+8z,\, Z= -6x-2y+8z,\hspace {5pt} x,y,z \in \mathbb R\big\}$
You can easily show that $Z=2Y-4X$.
So, $ImT = \big\{ (X,Y,2Y-4X): X,Y \in \mathbb R\big\}=\big\{X\cdot (1,0,-4)+Y\cdot (0,1,2):X,\, Y \in \mathbb R\big\}$.
That means $ImT=\big\langle (1,0,-4),\, (0,1,2) \big \rangle$, plus $(1,0,-4),\, (0,1,2)$ are linearly independent $\implies \dim ImT=2$.
Thus, we have $\dim \ker T+\dim ImT=\dim\mathbb R^3$.
Best Answer
First, plug the vectors into a matrix as columns (or rows). If you plugged them in as columns, find the column space. If you plugged them in as rows, then carry out row reduction to find the row space.
If the column space or the row space has dimension $<3$, then they cannot form a spanning set of $\mathbb{R}^3$. Note that, if you have $x$ vectors in $\mathbb{R}^n$, and $x>n$, then the set of vectors is never linearly independent, and thus cannot be a basis of $\mathbb{R}^n$.
However, if your vectors are a spanning set, then a subset of them will be a basis for $\mathbb{R}^n$.