Answer
$$v_{x}= -3.5\times10^{3} m/s$$
$$v_{y}=7 \times 10^{3} m/s$$
Work Step by Step
the magnetic force on the proton is determined via the relationship:
$$\vec{F}=q(\vec{v} \times \vec{B})$$
where the q is the charge of the proton and it is $+e$
$\vec{v}$ is the velocity of the proton = $v_{x}\vec{i}+v_{y}\vec{j}+2000\vec{k} $ $[m/s]$
and $\vec{B}$ is the magnetic field = $10\vec{i}-20\vec{j}+30\vec{k} $ $[mT]$
so , $$F_{x}=e(0.03v_{y}+40)=4\times 10^{-17} N$$
and $$F_{y}=e(20-0.03v_{x})=2\times 10^{-17} N$$
so $$v_{x}= -3.5\times10^{3} m/s$$
$$v_{y}=7 \times 10^{3} m/s$$
