jmc

Writing the expression as a quadratic in $x,$ we get $$3x^2-(4y+6z)x+\cdots .$$Thus, completing the square in $x,$ we get $$3{\left(x-\frac{2y+3z}{3}\right)}^2+\frac{32}{3}{y}^2-16yz+24z^2-8y-24z.$$We can then complete the square in $y,$ to get $$3{\left(x-\frac{2y+3z}{3}\right)}^2+\frac{32}{3}{\left(y-\frac{6z+3}{8}\right)}^2+18z^2-30z-\frac{3}{2}.$$Finally, completing the square in $z,$ we get $$3{\left(x-\frac{2y+3z}{3}\right)}^2+\frac{32}{3}{\left(y-\frac{6z+3}{8}\right)}^2+18{\left(z-\frac{5}{6}\right)}^2-14.$$Thus, the minimum value is $\boxed{-14},$ which occurs when $x-\frac{2y+3z}{3}=y-\frac{6z+3}{8}=z-\frac{5}{6}=0,$ or $x=\frac{3}{2},$ $y=1,$ and $z=\frac{5}{6}.$