firstly no $(a,b,c) =1$ let $a+b+c=k$ $\gcd(a, k-a)=\gcd(a,a-k) = \gcd(-k,a)=\gcd(k,a)>1$ same as the two other $\gcd(k,b)>1 ,\gcd(k,c)>1$ $\gcd(a,k) = x$ $\gcd(b,k) =y$ $\gcd(k,c) = z$ consider k has only one prime factor meaning $p \mid (a,b,c)$ contradiction if k has 2 prime factor $k = p^{a_1}*t^{a_2}$ $p^{a_1}*t^{a_2} = pa'+pb'+tc' $ $0 \equiv tc' \mod p$ meaning divisible by p showing (a,b,c) = p contradiction consider k has 3 prime factor $p^{a_1}*q^{a_2}*r^{a_3}=k$ $p^{a_1}*q^{a_2}*r^{a_3}= pa'+qb'+rc'$ now pick from the smallest $2*3*5$ $30 = 2a'+3b'+5c'$ $0 = b'+c' \mod 2$ but if b',c' both divisible by 2 meaning $gcd(a,b,c) =2$ $30 = 2(a')+3(2b''-1)+5(2c''-1)$ $38 = 2a'+6b"+10c"$ $a+3b"+5c"=19$ pick ,$c"=3,b"=1,a=1$ $c'=5,b'=1,a'=1$ $c = 25,b=3,a=2$ $a+b+c=30$ is the minimum