Effect of removing the product from equilibrium
First, let’s analyse the situation the graph presents. Obviously the reaction is (just restating for clarity):

$$\ce{PCl3 + Cl2 &lt;=&gt; PCl5}\tag{1}$$

We have already removed some $\ce{PCl5}$ from the mixture and are now looking at the self-reestablishing equilibrium. Obviously, for each molecule of $\ce{PCl5}$ we create we lose one molecule of $\ce{PCl3}$ and one of $\ce{Cl2}$ — this is the information we gain from the reaction coefficients (they are all $1$). This can then be expanded to the amount of substance reacting, namely $$\Delta n(\ce{PCl5}) = -\Delta n (\ce{PCl3}) = -\Delta n(\ce{Cl2})\tag{2}$$

Your graph uses concentration $c$ as $y$ axis. Concentration is defined as:

$$c = \frac nV\tag{3}$$

Since we can assume the volume $V$ to be constant, we see that concentration and amount are simply related by a constant factor; thus any differences in concentration will be analogous to differences in amount. Therefore, the labelled $\Delta x$ in the graph are indeed all identical as the represent the same concentration difference and thus the same difference in amount; and we have shown in equations $(1)$ and $(2)$ that the amounts are identical.



What about a case such as your hypothetical following case:

$$\ce{2A + B &lt;=&gt; 3 C}\tag{4}$$

Well, in general the same idea applies except that you must now take care of the different coefficients. Say we removed some $\ce{C}$ from the mixture. New $\ce{C}$ will be created and the creation of three molecules of $\ce{C}$ will use up two molecules of $\ce{A}$ and one of $\ce{B}$. Thus:


For each mole of $\ce{C}$ created we lose $\frac 13~\mathrm{mol}\ \ce{B}$
For each mole of $\ce{C}$ created we lose $\frac 23~\mathrm{mol}\ \ce{A}$


Therefore the relationship is:

$$\Delta n(\ce{C}) = -3\Delta n(\ce{B}) = -\frac 32\Delta n(\ce{A})\tag{5}$$