\((a \hat{i}+b \hat{j}+\hat{k}) \cdot(2 \hat{i}-3 \hat{j}+4 \hat{k})=0\)

\(2 a-3 b+4=0\)

On solving, \(2 a-3 b=-4\)

Also given

\(3 a+2 b=7\)

We get \(a =1, b =2\)

\(\frac{ a }{ b }=\frac{ x }{2} \Rightarrow x =\frac{2 a }{ b }=\frac{2 \times 1}{2}\)

\(\Rightarrow x =1\)