Answer
The meteor's speed when it hits the surface of the planet is 13.0 km/s
Work Step by Step
If the meteor's speed far away was zero, the kinetic energy it would gain before striking the planet's surface would be $\frac{1}{2}mv_e^2$. When the meteor's speed far away is non-zero, it will still gain the same amount of kinetic energy before it strikes the planet's surface.
We can find the meteor's speed when it hits the surface of the planet:
$KE = \frac{1}{2}mv_0^2+\frac{1}{2}mv_e^2$
$\frac{1}{2}mv^2 = \frac{1}{2}mv_0^2+\frac{1}{2}mv_e^2$
$v^2 = v_0^2+v_e^2$
$v = \sqrt{v_0^2+v_e^2}$
$v = \sqrt{(5.0~km/s)^2+(12.0~km/s)^2}$
$v = 13.0~km/s$
The meteor's speed when it hits the surface of the planet is 13.0 km/s.
