<s>[INST] Find three consecutive entries of a row of Pascal triangle that are in the ratio of 1 : 2 : 3 
Find three consecutive entries of a row of Pascal triangle that are in the ratio of 1 : 2 : 3

This means that:
$$\begin{align}
2\binom{n}{k} =\binom{n}{k+1}\\ 
3\binom{n}{k} =\binom{n}{k+2}\\ 
\end{align}$$
I simplefied these equations:
$$\begin{align}
2\binom{n}{k} =\binom{n}{k+1}\\ 
\end{align}$$
Is the same as: 
$$\begin{align}
2k+2 = n-k\\
3k+2 = n
\end{align}$$
And:
$$\begin{align}
3\binom{n}{k} =\binom{n}{k+2}\\ 
\end{align}$$
Is the same as: 
$$\begin{align}
3(k+1)(k+2) = (n-k-1)(n-k)
\end{align}$$
I do not know how I could go further so that I would end up with the value of n and k. This is because I can not simplify 3(k+1)(k+2) = (n-k-1)(n-k) enough.
How can I proceed so that I can get the value of n and k for:
$$\begin{align}
3(k+1)(k+2) = (n-k-1)(n-k) \\
3k+2 = n
\end{align}$$
 [/INST] Since $n=3k+2$, you can have
$$3(k+1)(k+2)=(\color{red}{3k+2}-k-1)(\color{red}{3k+2}-k)$$
Now solve this for $k$. (don't forget to check if each $k$ is sufficient.)
</s>