Answer
See the detailed answer below.
Work Step by Step
First, we need to find the wavelength of the three identical sources.
$$\lambda=\dfrac{v}{f}=\dfrac{340}{170}=\bf 2\;\rm m$$
$$\color{blue}{\bf [a]}$$
We know that the net amplitude at some point is determined by the interference of the given waves.
At the point you stand, the two sources 2 and 3 interact destructively since the path length difference of the two waves is half wavelength.
$$\Delta r_{3\rightarrow2}= r_3-r_2 = \sqrt{4^2+3^2}-4=1$$
Now the net amplitude depends only on the first source above
where we know that the amplitude of the wave from each speaker is $a$.
Therefore,
$$\boxed{A=a}$$
$$\color{blue}{\bf [b]}$$
If we canceled the middle source 2, the interference at the given point from 1 and 3 is perfectly constructive since $\Delta r_{1\rightarrow 3}=0$. This makes the net amplitude $2a$
So to have a net amplitude of $3a$, we need to move it to the left one half-wavelength. Then the three waves reach point P in phase.
$$\color{blue}{\bf [c]}$$
We know that the sound intensity is given by $$I=CA^2$$
So, when the amplitude is maximum, $A=3a$
$$I_{max}=C(3a)^2=9Ca^2$$
and from a single source,
$$I_{single}=Ca^2$$
Hence,
$$\boxed{I_{max}=9I_{single}}$$
