That was a close call!

At the instant of closest approach the particle velocity is perpendicular to the radius vector joining the center of the Earth to the particle. Thus applying law of conservation of angular momentum, we get

$mv_0a = mv(R_e+b).$

where $v$ is the speed of the particle at the instant of closest approach.

For the sake of calculations we'll denote $R_e + b$ as $r$ .

So, from the above equation

$v = \dfrac{a v_0}{r} \qquad ...(1)$

Also applying conservation of mechanical energy, we get

$\dfrac 12 m {v_0}^2 = \dfrac 12 m v^2 + E \qquad ...(2)$

where $E$ denotes the field intensity of Earth at distance $r$ from the center.

Now,

$E = - \dfrac{GM_e m}{r} \qquad ...(3)$

Substituting the value for $v$ and $E$ from equations $(1)$ and $(3)$ in equation $(2)$ , we get

$\begin{aligned} \dfrac 12 m {v_0}^2 &= \dfrac 12 m \dfrac{a^2 \ {v_0}^2}{r^2} - \dfrac{GM_e m}{r} \\ \implies \dfrac 12 \left( \dfrac{a^2 \ {v_0}^2}{r^2} - {v_0}^2 \right) &= \dfrac{GM_e}{r} \\ \implies \dfrac{{v_0}^2}{2} \left( \dfrac{a^2 - r^2}{r^2} \right) &= \dfrac{GM_e}{r} \\ \implies {v_0}^2 a^2 - {v_0}^2 r^2 &= 2GM_e r \\ \implies {v_0}^2 r^2 + 2GM_e r - {v_0}^2 a^2 &= 0. \end{aligned}$

Which is a quadratic in $r$ . Solving for $r$ , we get

$\begin{aligned} r &= \dfrac{-2GM_e + \sqrt{4G^2{M_e}^2 + 4 {v_0}^4 a^2}}{2{v_0}^2} \qquad \qquad \small \color{#3D99F6}{\text{Neglecting negative value}} \\ &= -\dfrac{GM_e}{{v_0}^2} + \dfrac{GM_e}{{v_0}^2} \sqrt{ 1 + \dfrac{{v_0}^4 a^2}{G^2 {M_e}^2}} \\ &= \dfrac{GM_e}{{(v_0)}^2} \left[ \sqrt{1 + {\left( \dfrac{a \ {v_0}^{2}}{\ GM_e} \right)}^2} -1 \right]. \end{aligned}$

As $r = b + R_e$ , therefore

$b = \dfrac{GM_e}{{(v_0)}^2} \left[ \sqrt{1 + {\left( \dfrac{a \ {(v_0)}^{2}}{\ GM_e} \right)}^2} -1 \right] - R_e.$

which gives

$x=1 \\ y=2 \\ z=1 \\ \implies x+y+z = \boxed{4}.$

A really nice problem, that I decided to post my handwritten solution to it, even though it shares a similar approach discussed in past solutions here. Enjoy.