The resultant of two forces $$P$$ N and 3 N is a force of 7 N. If the direction of 3 N force were reversed, the resultant would be $$\sqrt{19}$$ N. The value of $$P$$ is

Let the two forces be represented by the vectors $$\vec{P}$$ (magnitude $$P$$ N) and $$\vec{Q}$$ (magnitude $$3$$ N).
Let the angle between them be $$\theta$$.

Case 1: Both forces act in their given directions.
The magnitude of their resultant $$\vec{R}$$ is given by the law of vector addition:

$$R=\sqrt{P^{2}+3^{2}+2P\cdot3\cos\theta}$$

We are told that $$R = 7$$ N, hence

$$P^{2}+9+6P\cos\theta = 7^{2}=49$$

$$\Rightarrow \; P^{2}+6P\cos\theta = 40 \quad \;-(1)$$

Case 2: The 3 N force is reversed, i.e. replaced by $$-\vec{Q}$$. The angle between $$\vec{P}$$ and $$-\vec{Q}$$ is still $$\theta$$, but the cosine term changes sign. The new resultant $$\vec{R}'$$ therefore has magnitude

$$R'=\sqrt{P^{2}+3^{2}-2P\cdot3\cos\theta}$$

Given $$R'=\sqrt{19}$$ N, we get

$$P^{2}+9-6P\cos\theta = (\sqrt{19})^{2}=19$$

$$\Rightarrow \; P^{2}-6P\cos\theta = 10 \quad \;-(2)$$

Now add equations (1) and (2):

$$\big(P^{2}+6P\cos\theta\big)+\big(P^{2}-6P\cos\theta\big)=40+10$$

$$2P^{2}=50 \quad\Rightarrow\quad P^{2}=25$$

Since a force magnitude is positive, $$P=\sqrt{25}=5$$ N.

Therefore, the required magnitude of the first force is 5 N.
Option A which is: 5 N