Because the role of a,y,z is the same so we suppose \(x\le y\le z\)

\(2^z< 2336\Rightarrow z\le11\)

We have:\(2^x+2^y+2^z\le3.2^z\Leftrightarrow2336\le3.2^z\)\(\Leftrightarrow2^z\ge\dfrac{2336}{3}\Rightarrow z\ge10\)

So z\(\in\left\{10,11\right\}\)

With z=10 then \(2^x+2^y=2336-2^{10}=1312\)

    \(2^x+2^y\le2.2^y\Rightarrow2.2^y\ge1312\Rightarrow2^y\ge656\Rightarrow y\ge10\)

  \(2^y< 1312\Rightarrow y\le10\)

 So y=10 so \(2^x=2336-2^{10}-2^{10}=288\)(not satisfy)

With z=11 then \(2^x+2^y=288\)

 \(2^y< 288\Rightarrow y\le8\)

 \(2^y.2\ge2^x+2^y=288\Rightarrow2^y\ge144\Rightarrow y\ge8\)

So y=8 then \(2^x=288-2^8=32\Rightarrow x=5\)

Answer :(x,y,z) is the permutation of \(\left(5,8,11\right)\)

oh;the first row :x,y,z not a,y,z