jmc

By Cauchy-Schwarz, $$\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{6}\right)(2a^2+3b^2+6c^2)\ge (a+b+c{)}^2=1,$$so $2a^2+3b^2+6c^2\ge 1.$

Equality occurs when $4a^2=9b^2=6c^2$ and $a+b+c=1.$ We can solve to get $a=\frac{1}{2},$ $b=\frac{1}{3},$ and $c=\frac{1}{6},$ so the minimum value is $\boxed{1}.$