How many solution?

This was a straightforward complete the square problem.

$\begin{aligned} x^2+y^2+z^2+4 &= xy+3y+2z\\ x^2-xy+\frac{1}{4}y^2 + \frac{3}{4} y^2 - 3y + 3 + z^2 - 2z + 1 &= 0\\ \left ( x - \frac{1}{2}y \right )^2 + \frac{3}{4}(y-2)^2 + (z-1)^2 &= 0 \end{aligned}$

Since squares are nonnegative, it follows that $(x,y,z)=(1,2,1)$ , and this is the only solution. Thus, there is one solution only.