Launching a Plaintext Attack against Affine Cipher

Update 2

Being new to the world of Stack Exchange I did not realize that there exists a site solely devoted to cryptography. In light of this, I hope someone could help me migrate this question to the appropriate site. Unless, of course, someone could answer the question here instead. I am not getting anywhere with this question.

Background

Last year, a question concerning plaintext attacks was posted on on this website. I have no problem to see how to solve it when we are given two ciphertexts and $$(c_1,c_2)$$ and their corresponding plaintexts $$(m_1,m_2)$$. But, when I am to deal with the situation where $$p$$ is unknown it gets complicated. In this instance, I have three pairs of ciphertexts and plaintexts - $$c_i \not =c_j$$ for $$i,j \in \{1,2,3\}$$ and $$m_i \not = m_j$$ for $$i,j \in \{1,2,3\}$$. This differs then from the previous question in the sense that I cannot use the method that fkraiem provided given two of the ciphertexts are equal.

To find $$p$$:

You have $$k_1m_1+k_2 \equiv k_1m_2+k_2 \equiv c_2 \pmod p$$ so $$k_1(m_1-m_2) \equiv 0 \pmod p$$. This means that either $$k_1$$ or $$m_1-m_2$$ is a multiple of $$p$$ (this is where the fact that $$p$$ is prime comes in). $$k_1$$ can't be a multiple of $$p$$, because otherwise the encryption function is constant, which is absurd, so $$m_1-m_2$$ is a multiple of $$p$$.

You can now try to find $$k_1,k_2$$ using each prime divisor of $$m_1-m_2$$ as modulus until you find one which works for the two pairs $$((m_1,c_2),(m_3,c_3))$$ and $$((m_2,c_2),(m_3,c_3))$$.

Thus my question is this: How would one go about determine $$p$$ if the ciphertexts are all different?

Update

For example, suppose $$(c_1,c_2,c_3)=(13,19,3)$$ and $$(m_1,m_2,m_3)=(5,10,23)$$. How do I go about to determine $$p$$?

You know from the affine equations that there are $t_i$ such that $5k_1 + k_2 = 13 + t_1p, 10k_1 +k_2 = 19+t_2p$ and $23k_1+k_2=3+t_3p$. The first 2 subtracted tells us that $5k_1 -6$ is a multiple of $p$, and the third minus the first tells us $18k_1+10$ is a multiple of $p$ as well.
Multiply $5k_1 -6$ by $18$ and $18k_1+10$ by $5$ and we still have multiples of $p$. Subtract and we know $158$ is a multiple of $p$.