smc

We examine each of the conditions. The first condition is false. A simple counterexample is $a=3$ and $b=5$. The corresponding value of $c$ is $17\cdot 15=255$. Clearly, $\gcd (3,14)=1$ and $\gcd (5,15)=5$, so condition $I$ would imply that $\gcd (c,210)=1.$ However, $\gcd (255,210)$ is clearly not $1$ (they share a common factor of $15$). Obviously, condition $I$ is false, so we can rule out choices $A,B,$ and $C$. We are now deciding between the two answer choices $D$ and $E$. What differs between them is the validity of condition $II$, so it suffices to simply check $II$. We look at statement $II$'s contrapositive to prove it. The contrapositive states that if $\gcd (a,14)\ne 1$ and $\gcd (b,15)\ne 1$, then $\gcd (c,210)\ne 1.$ In other words, if $a$ shares some common factor that is not $1$ with $14$ and $b$ shares some common factor that is not $1$ with $15$, then $c$ also shares a common factor with $210$. Let's say that $a=a^{\prime }\cdot n$, where $a^{\prime }$ is a factor of $14$ not equal to $1$. (So $a^{\prime }$ is the common factor.) We can rewrite the given equation as $15a+14b=c⟹15(a^{\prime }n)+14b=c.$ We can express $14$ as $a^{\prime }\cdot n^{\prime }$, for some positive integer $n^{\prime }$ (this $n^{\prime }$ can be $1$). We can factor $a^{\prime }$ out to get $a^{\prime }(15n+14n^{\prime })=c.$ We know that all values in this equation are integers, so $c$ must be divisible by $a^{\prime }$. Since $a^{\prime }$ is a factor of $14$, $a^{\prime }$ must also be a factor of $210$, a multiple of $14$. Therefore, we know that $c$ shares a common factor with $210$ (which is $a^{\prime }$), so $\gcd (c,210)\ne 1$. This is what $II$ states, so therefore $II$ is true. Thus, our answer is $\boxed{\text{II and III only}}.$