Let us write $v_i = (x_i, y_i)$ for $1 \le i \le 3$. If $v_1$, $v_2$, $v_3$ are affine independent (i. e. there is no line containing all three of them), we can compute the (uniquely determined) affine function $f\colon \mathbb R^2 \to \mathbb R$ with $f(v_i) = D_i$ for $1 \le i \le 3$ an use it to interpolate other values. $f$ can be computed as follows:
As an affine function, $f$ has the form $f(x,y) = ax+by+c$, $a$, $b$, $c$ to be determined, so we plug in what we know giving a linear system of equations
ax_1 + by_1 + c &= D_1\\
ax_2 + by_2 + c &= D_2\\
ax_3 + by_3 + c &= D_3
Under our assumptions on the $v_i$, the equations have a unique solution $(a,b,c)$.
So I'll try to add something more about the "why". You should think about the one-dimensional case, that is: given points $x_1$, $x_2\in \mathbb R$, and same values $D_1$, $D_2$, we try to inter- or extrapolate the value at some other point $x$. To do this, we find the line connecting the points $(x_1, D_1)$ and $(x_2, D_2)$ in the plane and take it as a graph of a affine-linear function $f\colon\mathbb R \to \mathbb R$, and interpolate at $x$ by computing the value $f(x)$.
We follow this well known (I hope so) procedure in our two-dimensional case. The analogon of the line in 1D is a plane, and indeed: The graph of an affine-linear function $f\colon \mathbb R^2 \to \mathbb R$ is a plane. The equation of such a plane is $f(x,y) = ax + by + c$, that is to dertermine our plane we need to compute $a$, $b$ and $c$. As we want $f$ to interpolate the given values, we need $f(x_i, y_i) = D_i$, $1 \le i \le 3$. We compute them as above (solving the linear system $f(x_i, y_i) = D_i$). Then we just interpolate by evaluating $f$ (as we did in 1D).