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

Since the moment of $$\vec F$$ about $$A$$ is zero, the line of action of $$\vec F$$ passes through $$A$$.

Let the perpendicular distances of the line of action of $$\vec F$$ from $$B$$ and $$C$$ be $$d_B$$ and $$d_C$$ respectively.

Given,

$$Fd_B=9$$

and

$$Fd_C=16$$

Therefore,

$$d_B=\frac9F,\qquad d_C=\frac{16}{F}$$

Since the line of action passes through $$A$$, the distances of $$B$$ and $$C$$ from this line satisfy

$$BC^2=d_B^2+d_C^2$$

because $$\angle A=90^\circ$$.

Now,

$$BC=a=5$$

Hence,

$$\left(\frac9F\right)^2+\left(\frac{16}{F}\right)^2=5^2$$

$$\frac{81+256}{F^2}=25$$

$$\frac{337}{F^2}=25$$

$$F^2=\frac{337}{25}$$

This is not among the options, so instead use the fact that

$$AB=3,\qquad AC=4,\qquad BC=5$$

Let the line of action through $$A$$ make angle $$\theta$$ with $$AB$$.

Then,

$$d_B=AB\sin\theta=3\sin\theta$$

and

$$d_C=AC\cos\theta=4\cos\theta$$

Using the moments,

$$3F\sin\theta=9$$

$$4F\cos\theta=16$$

Hence,

$$F\sin\theta=3$$

$$F\cos\theta=4$$

Squaring and adding,

$$F^2(\sin^2\theta+\cos^2\theta)=3^2+4^2$$

$$F^2=25$$

$$F=5$$

Hence,

$$\boxed{5\text{ N}}$$