Let $x_1,$ $x_2,$ $x_3$ be positive real numbers such that $x_1 + 2x_2 + 3x_3 = 60.$  Find the smallest possible value of
\[x_1^2 + x_2^2 + x_3^2.\]
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 $\boxed{\frac{1800}{7}}.$