In a right angle $△ABC,$ $\angle A=90^{\circ }$ and sides $a,b,c$ are respectively, $5\text{cm},4\text{cm}$ and $3\text{cm}.$ If a force $\vec{F}$ has moments $0,9$ and $16$ in $\text{N cm}$ units respectively about vertices $A,B$ and $C,$ then magnitude of $\vec{F}$ is

The moment of a force about a point is the product of the force and the perpendicular distance from the point to the line of action of the force.
Moment about $A=0\Rightarrow$ the line of action of $\vec{F}$ passes through $A$.
Let the force $\vec{F}$ make an angle $\theta$ with side $AB$ (which is $c=3\text{cm}$).
Moment about $B=F\times (c\sin \theta )=F\times (3\sin \theta )=9\Rightarrow F\sin \theta =3$.
Side $AC$ is perpendicular to $AB$ (since $\angle A=90^{\circ }$). The perpendicular distance from $C$ to the line of action of $\vec{F}$ passing through $A$ is $b\cos \theta =4\cos \theta$.
Moment about $C=F\times (4\cos \theta )=16\Rightarrow F\cos \theta =4$.
Squaring and adding:
$F^2({\sin }^2\theta +{\cos }^2\theta )=3^2+4^2$
$F^2=9+16=25\Rightarrow F=5$.