Lemma: Let $R$ be an integral domain, $f,g\in R[X]$. The following is a necessary and sufficient condition for $f$ and $g$ not to be coprime: There exist $s,t\in R[X]$ such that $1\le \deg s<\deg g$ and $1\le \deg t<\deg f$, while $$fs+gt=0.$$ Proof: (necessary:) Suppose $f$ and $g$ share a factor $d$ with $\deg d >1$. Take $s=\frac gd$ and $t=-\frac td$.
(sufficient:) Suppose $s$ and $t$ exist as described, then $fs=-gt$. If $f$ and $g$ are coprime, then $f\mid t$ which contradicts $\deg t<\deg f$. Hence, $f$ and $g$ are not coprime. Q.E.D.
We can imitate the proof of the lemma to show that for $f,g,h\in R[X]$ not to be coprime, it is necessary that there exist $s,t,u\in R[X]$ with $$ftu+gsu+hst=0$$ and \begin{align*} 1&\le \deg s < \deg f\\ 1&\le \deg t < \deg g\\ 1&\le \deg u < \deg h. \end{align*}
Question: Is the above condition sufficient for $f,g,h$ not to be coprime?
The condition is not sufficient. Take for example \begin{align*} f&=X^2\\ g&=X^2+1\\ h&=-2X^2-1\\ s&=t=u=X, \end{align*} then clearly \begin{align*} 1&\le \deg s < \deg f\\ 1&\le \deg t < \deg g\\ 1&\le \deg u < \deg h \end{align*} and $$ftu + gsu + hst = X^2(x^2+X^2+1-2X^2-1)=0,$$ by $f,g,h$ are coprime over any $R[X]$.