The problem is:
The points $(2,-1,-2)$, $(1,3,12)$, and $(4,2,3)$ lie on a unique plane. Where does this plane cross the $z$-axis?
I can easily solve this problem by calculus and cross product: the equation of a plane is $$-22x+33x-11z=-55.$$ Hence, this plane crosses the $z$-axis at $z=5$. However, the problem requires to be sold by the system of linear equations transformed into matrix. Even vectors have not been introduces yet. Just a matrix in echelon form and a back substitution. Hence, when I am writing three equations: $$2a-b-2c=d$$ $$a+3b+12c=d$$ $$4a+2b+3c=d$$ I end up with 4 variables to find. What am I missing? Is there one more linear equation possible to add?
By Gaussian elimination,
$$\begin{cases}2a-b-2c=d,\\ a+3b+12c=d,\\ 4a+2b+3c=d\end{cases}$$
$$\begin{cases}\dfrac72b+13c=\dfrac d2,\\ 4b+7c=-d\end{cases}$$
$$-\frac{55}2c=-\frac{11}2d$$ and the requested point satisfies
$$a\,0+b\,0+c\,z=d.$$
So $z=5.$