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:
$$\begin{align}
3(k+1)(k+2) = (n-k-1)(n-k) \\
3k+2 = n
\end{align}$$

Alternatively, coming from this question, dividing the consecutive binomial coefficients yields
$$\frac21=\frac{\tbinom{n}{k+1}}{\tbinom nk}=\frac{n-k}{k+1}\qquad\text{ and }\qquad
\frac32=\frac{\tbinom{n}{k+2}}{\tbinom{n}{k+1}}=\frac{n-k-1}{k+2},$$
and solving both for $n$ yields $n=3k+2$ and $n=\tfrac52k+4$, respectively, so $k=4$ and $n=14$.
