When forces$F_1, F_2, F_3$  are acting on a particle of mass $m$ such that $F_2$ and $F_3$ are mutually perpendicular, then the particle remains stationary. If the force $F_1$ is now removed then the acceleration of the particle is 

When $F_1$, $F_2$ and $F_3$ are acting on a particle then the particle remains stationary. This means that the resultant of $F_1{F}_2$ and $F_3$ is zero. When $F_1$ is removed, $F_2$ and $F_3$ will remain. But the resultant of $F_2$ and $F_3$ should be equal and opposite to $F_1$.

i.e. $\left|\stackrel{\rightharpoonup }{F_2}+\stackrel{\rightharpoonup }{F_3}\right|=\left|\stackrel{\rightharpoonup }{F_1}\right|$

$\therefore a=\frac{\left|\stackrel{\rightharpoonup }{F_2}+\stackrel{\rightharpoonup }{F_3}\right|}{m}\Rightarrow a=\frac{F_1}{m}$