Answer
$-(1.002\times10^{-19}\hat i+6.68\times10^{-20}\hat j)\;kg.m/s$
Work Step by Step
The linear momentum of Particle 1:
$\vec{P_1}=(16.7 \times10^{-27}\times6.00 \times10^6\;kg.m/s)\hat i=(1.002\times10^{-19}\;kg.m/s)\hat i$
The linear momentum of Particle 2:
$\vec{P_2}=(8.35 \times10^{-27}\times8.00 \times10^6\;kg.m/s)\hat j=(6.68\times10^{-20}\;kg.m/s)\hat j$
If $\vec{P_3}$ be the linear momentum of the Particle 3, then according to the theory of conservation of linear momentum
$\vec{P_1}+\vec{P_2}+\vec{P_3}=0$
or, $\vec{P_3}=-(\vec{P_1}+\vec{P_2})$
or, $\vec{P_3}=-(1.002\times10^{-19}\hat i+6.68\times10^{-20}\hat j)\;kg.m/s$
