10. Let $x+y+z=1, p=2 x^{2}+3 y^{2}+z^{2}$, find $x$ when $p$ takes its minimum value.
10. $\frac{3}{11}$.
$$
p=2 x^{2}+3 y^{2}+(1-x-y)^{2}=3 x^{2}+(2 y-2) x+4 y^{2}-2 y+1=3\left(x+\frac{y-1}{3}\right)^{2}+
$$
$\frac{11}{3}\left(y-\frac{2}{11}\right)^{2}+\frac{1}{11}, p_{\min }=\frac{6}{11}$, at this point $x+\frac{y-1}{3}=0$, and $y=\frac{2}{11}$, solving gives $x=\frac{3}{11}$.