jmc

By Cauchy-Schwarz, $$(1+4+9)(x_1^2+x_2^2+x_3^2)\ge (x_1+2x_2+3x_3{)}^2=60^2,$$so $x_1^2+x_2^2+x_3^2\ge \frac{3600}{14}=\frac{1800}{7}.$

Equality occurs when $x_1=\frac{x_2}{2}=\frac{x_3}{3}$ and $x_1+2x_2+3x_3=60.$ We can solve, to find $x_1=\frac{30}{7},$ $x_2=\frac{60}{7},$ and $x_3=\frac{90}{7}.$ Hence, the smallest possible value is $\boxed{\frac{1800}{7}}.$