Consider $a,b,c\in\mathbb{R}-\{0\}.$ A real valued function $f$ is defined as $f(x,y,z)=2^y\cdot3^z\cdot a^x\cdot b^y\cdot c^z+2^z\cdot3^x\cdot a^y\cdot b^z\cdot c^x+2^x\cdot3^y\cdot a^z\cdot b^x\cdot c^y$, where $x,y,z\in\mathbb{Z}$. Given that $f(1,0,0)=4$ and $f(2,0,0)=6$ then
The possible value of $[a]$ $($where $[.]$ denotes the greatest integer function$)$ can be: Answer given is $0,1,2$
The integral values of $\frac{1}{b}$ can be: Answer given is $1,2,3$
If all possible values of $c$ satisfying given condition is $[p,q],$ then $4q-3p$ is equal to : Answer given is $2$
The maximum value of $[a]^2+[b]^2-[c]$ $($where $[.]$ denotes the greatest integer function$)$ is: Answer given is $5$
My work: We can form two equations: $$a+2b+3c=4$$ and $$a^2+4b^2+9c^2=6$$ After this, I wasn't able to do anything. Then I checked if there are solutions to this on Wolfram Alpha and I got two set of real solutions. $$a=\frac{2}{3},\:\:b=\frac{5}{6},\:\:c=\frac{5}{9}$$and $$a=2,\:\:b=\frac{1}{2},\:\:c=\frac{1}{3}$$ In both the cases, the value of $[a]$ is either $2$ or $0$. It can't be $1$. Likewise, there are many loopholes in the other given answers also. So this can mean two things. Either the printed answers are wrong, or WA is not giving full solutions $($ though I highly doubt this case$)$.
Also, how can one get two or more set of real solutions$?$ Do we have to form a parametric equation or do we have to apply some advance algebra$?$
I actually require the method by which we can get values of $a,b,c$ and then I can review the printed answers.
Any help is greatly appreciated.
Try to form quadratica in all variables then apply the condition of real roots. You will get values of all variables having upper and lower bound.