Example 4 (1992 "Friendship Cup" International Mathematics Invitational for Grade 8) Find three real numbers $x, y, z$, such that they simultaneously satisfy the following equations: $2 x+3 y+z=13,4 x^{2}+9 y^{2}+z^{2}-2 x+15 y+$ $3 z=82$.
Analysis: By adding and completing the square of the two equations, we get $(2 x)^{2}+$ $(3 y+3)^{2}+(z+2)^{2}=108$. Write the first equation as $2 x$ $+(3 y+3)+(z+2)=18$. From this and the Cauchy-Schwarz inequality, we have $108 \cdot 3=\left[(2 x)^{2}+(3 y+3)^{2}+(z+2)^{2}\right]\left(1^{2}\right.$ $\left.+1^{2}+1^{2}\right) \geqslant[(2 x)+(3 y+3)+(z+2)]^{2}=18^{2}$.

The equality in the above equation holds if and only if $2 x=3 y+3=z+$ 2. Also, from $2 x+(3 y+3)+(z+2)=18$ we get $x=3, y=$ $1, z=4$.