The equation is
$$(y_0z_1-y_1^2) \sum\limits_{i=0}^2 c_i x_i=x_0x_1z_1+y_0\left(\sum\limits_{i=0}^2 c_i y_i\right)\left(c_0y_1+\sum\limits_{i=1}^2 c_i z_i\right) -y_1x_0\left(c_0y_1+\sum\limits_{i=1}^2 c_i z_i\right) -y_1x_2\left(\sum\limits_{i=0}^2 c_i y_i\right)$$
The unknown are $c_0,c_1,c_2$, the other variables are fixed on the real line under the following assumptions: $y_0\geq 0, z_1 \geq 0, y_0z_1-y_1^2 \neq 0$.
Is there a real solution to this equation under those general assumptions ? It looks like the equation corresponds to finding the real root(s) of a polynomial of degree $2$ in $c_0,c_1,c_2$. Is this possible ? I only need to know that a real solution exists, not necessarily compute it.
The challenge is hidden in the square roots we take to arrive at a solution: for example if we fix $c_0=0=c_1$, we get a quadratic equation for $c_2$, but then nothing guarantees that $c_2$ is real.
If there are no real solutions under those assumptions, can you find what extra (minimal) assumptions should the variables satisfy to get a real solution ?
Under the given assumption, the equation has no real solutions iff $$y_0=x_0z_1-x_1 y_1+x_2 y_1=x_0z_2- x_2 y_1+x_2y_2=0\mbox{ and }x_0x_1z_1\ne 0.$$
Let us show this. I did routine calculations with Mathcad, so we skip them.
Impose first a restriction $c_2=0$. Then the initial equation becomes a quadratic equation with respect to $c_1$ whose discriminant is a quadratic polynomial with respect to $c_0$ with the highest coefficient $y_0^2(y_0z_1-y_1^2)^2$. So if $y_0\ne 0$ then for sufficiently big $c_0$ the discriminant is non-negative and so the initial equation has a real solution.
If $y_0=0$ then $y_1\ne 0$ and the initial equation reduces to a linear equation
$$(x_0z_1-x_1 y_1+x_2 y_1)c_1+(x_0z_2- x_2 y_1+x_2y_2)c_2= x_0x_1z_1/y_1.$$