for minimality $(a,b,c)=1$ and so $b=x^2,c=y^2,a=xy$ where $x \neq y$ $k=2x^2+3y^2-xy=\frac{ (4x-y)^2+23y^2}{8}=f(x,y)$ If $y \geq 2$ then $f(x,y) \geq \frac{23}{2}$ $f(2,1)=9$ and $f(x,1) \geq f(2,1)$ so $k \geq 9$ and $k=9$ for $b=4,c=1,a=2$

Assume $a=1 \Longrightarrow b=c=1$, contradiction! Now by AM-GM $k=2b+3c-a \ge 2\sqrt{6}\sqrt{bc}-a = (2\sqrt{6}-1)a \ge 2(2\sqrt{6}-1) >7$. Hence $k \ge 8$. Assume now that $k=8$, then $2b+3c-8=a \Longrightarrow (2b+3c-8)^2=a^2 \Longrightarrow 64 = (2b-4)^2+(3c-8)^2+11a^2 \ge 11a^2$ This means we must have $a=2$, but then $(b,c)=(4,1)$ or $(1,4)$, yielding $k=9$ or $k=12$. Hence we have $k>8$ and the minimal value of $k=9$ can be achieved at $(a,b,c)=(2,4,1)$,

First by AM-GM we have $2b+3c-a>4\sqrt(bc)+c-a=3a+c$. The smallest value of $a$ that has distinct $a,b,c$ is $a=2$ with $(b,c)=(1,4),(4,1)$. Testing these $(2,4,1)$ gives a minimal $k$ of $9$. If $3a+c<k<9$ then $a<3$ which we already checked.