I assume you're trying to show that $\frac{d(d+3)}{2}$ points determine a degree $d$ plane curve.
First, for $d = 1$, you're already most of the way there. Here is one way to argue. (I'm going to assume everything is in projective space. It's not that much more difficult in the affine case.) Suppose we have two distinct points $[X_1:Y_1:Z_1]$ and $[X_2:Y_2:Z_2]$. Then your system of equations can be written
\begin{equation}
\underbrace{\begin{pmatrix} X_1 & Y_1 & Z_1 \\ X_2 & Y_2 & Z_2 \end{pmatrix}}_{T} \begin{pmatrix} a \\ b \\ c \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix}
\end{equation}
The space of solutions to this linear system is itself a vector space, namely $\ker T$, and its dimension can be found using the rank-nullity theorem. The rank of $T$ is 2 since the two points are distinct. Therefore $\dim \ker T = 3 - 2 = 1$. So the space of solutions $\{(a,b,c)\}$ is a one-dimensional vector space, and therefore all such solutions are multiples of each other.
For larger values of $d$, I am going to take a different approach than trying to explicitly find the degree $d$ curve passing through the given points. Since you only asked for ideas, I will leave you to fill in the details. A plane curve is essentially a nonzero homogeneous polynomial $f \in k[X,Y,Z]$ thought of as modulo scaling. The space of all such polynomials is itself a projective space with coordinates given by its coefficients. For example, when $d = 1$, the line $aX + bY + cZ = 0$ corresponds to the point $[a:b:c] \in \mathbb{P}^2$.
What is the dimension of this "dual" projective space that parametrizes degree $d$ curves?
The answer is $\frac{d(d+3)}{2}$. (Do combinatorics and count the number of degree $d$ monomials. Don't forget to subtract 1!)
Now consider what happens when I require the curve to pass through a given point $[X_1:Y_1:Z_1]$. Then I need the coefficients of the degree $d$ curve $[a_0: \cdots : a_{d(d+3)/2}]$ to satisfy the linear equation
\begin{equation}
X_1^d a_0 + X_1^{d-1} Y_1 a_1 + \cdots + Z_1^d a_{d(d+3)/2} = 0.
\end{equation}
This defines a hyperplane in $\mathbb{P}^{d(d+3)/2}$, which cuts down the dimension by 1. (Essentially the problem now becomes a linear algebra one.) Similarly, imposing the requirement that the curve passes through the other points means intersecting the space of degree $d$ curves with more hyperplanes, one for each point.
It is not difficult to show that with each additional hyperplane, the dimension of the space of degree $d$ curves remaining either remains the same (if the hyperplane contains the previous intersection) or goes down by one. After cutting the space by $\frac{d(d+3)}{2}$ hyperplanes, we are left with at worse a 0-dimensional space, i.e., a point. This point corresponds to the unique degree $d$ curve that passes through the $\frac{d(d+3)}{2}$ points that you are looking for.
But in general this need not be unique. For example, take $d = 2$ and suppose the five points are collinear. Then any conic that is two lines with one of the lines containing the five points passes through all five points. This means, in terms of the above discussion, that some of the hyperplanes defined by the points don't reduce the dimension. On the other hand, if they do, then we say that the points are in general position.
I'm not quite sure about your background, so some of this might need to be thought through carefully if this is the first time you're seeing this. Write back if anything is unclear.
