Since: \[6=(a-b)^2+(b-c)^2+(c-a)^2\ge \frac{1}{2}(a-c)^2+(c-a)^2=\frac{3}{2}(a-c)^2\] it follows that $|a-c|\le 2$. Then: \[\begin{aligned} (a+5)^2+(b-2)^2+(c-9)^2 &=(a-1)^2+(b-2)^2+(c-3)^2+12(a-c)+96\\ &\ge 12(a-c)+96\\ &\ge 72\\ \end{aligned}\] Equality occurs when $(a,b,c)=(1,2,3)$.

i have another one: Let $a=c+\alpha $, $b=c+\beta $ where $ \alpha , \beta \in R$. Us given that $ \alpha^2+\beta^2-\alpha \cdot \beta =3$ and $ -2 \le \alpha \le 2 $. \[ (a+5)^2+(b-2)^2+(c-9)^2=3c^2+2c(\alpha +\beta -6)+\alpha^2+\beta^2+10\alpha-4\beta +110 \] and this is parobola there exists minimun value $ min =100+14\alpha \ge 72. $

My solution: Considering the quadratic equation in $b$ we have $\Delta_b=12-3(a-c)^2\Longrightarrow -2\le a-c\le 2$. Denote $t=a+b+c$ so $$(a+5)^2+(b-2)^2+(c-9)^2=a^2+b^2+c^2+10a-4b-18c+110=$$$$2+\frac{t^2}{3}-4(a+b+c)+14(a-c)+110=$$$$\frac{1}{3}(t-6)^2+14(a-c)+100\ge100-28=72.$$The equality hold when $a-c=-2\Longrightarrow \Delta_b=0\Longrightarrow 2b=a+c$ and $a+b+c=6$, so $a=1$, $b=2$, $c=3$.

We find the solution $$ a=1,b=2,c=3 $$let $$ F(a,b,c)=a^2+b^2+c^2-ab-bc-ac $$Now,we prove that if either a,b or c increases,then F(a,b,c) increases. Assuming a increses with the value x,then we have $$ F(a+x,b,c)>F(a,b,c) $$which is equivalent with $$ (a+x)(a+x-b-c)> a(a-b-c) $$which is true,because x>0 Therefore,the only solution is $$ a=1,b=2,c=3 $$