smc

Given that all the answer choices and coefficients are integers, we hope that $P(x)$ has positive integer coefficients. Throughout this solution, we will express all polynomials in base $x$. E.g. $x^2+x+1=111_x$. We are given: $$111a+12=101b+21=P(x).$$ We add $111$ and $101$ to each side and balance respectively: $$111(a-1)+123=101(b-1)+122=P(x).$$ We make the unit's digits equal: $$111(a-1)+123=101(b-2)+223=P(x).$$ We now notice that: $$111(a-11)+1233=101(b-12)+1233=P(x).$$ Therefore $a=11_x=x+1$, $b=12_x=x+2$, and $P(x)=1233_x=x^3+2x^2+3x+3$. $3$ is the minimal degree of $P(x)$ since there is no way to influence the $x$'s digit in $101b+21$ when $b$ is an integer. The desired sum is $1^2+2^2+3^2+3^2=\boxed{23}$ P.S. The four computational steps can be deduced through quick experimentation.