1. Let the two vectors be $\vec{A}=a\hat{i}+b\hat{j}+\hat{k}$ and $\vec{B}=2\hat{i}-3\hat{j}+4\hat{k}.$

2. Since the vectors are perpendicular, their dot product must be zero: $\vec{A}\cdot\vec{B}=0.$

3. Compute the dot product: $\vec{A}\cdot\vec{B}=a(2)+b(-3)+1(4)=0$ $2a-3b+4=0.$

4. We are also given: $3a+2b=7.$

5. Solve the system of equations: $2a-3b=-4 \qquad ...(1)$ $3a+2b=7 \qquad ...(2)$

6. Eliminate variables. Multiply (1) by $2$ and (2) by $3$: $4a-6b=-8$ $9a+6b=21$

   Adding, $13a=13 \Rightarrow a=1.$

7. Substitute into (2): $3(1)+2b=7$ $3+2b=7$ $2b=4 \Rightarrow b=2.$

8. Therefore, $\frac{a}{b}=\frac{1}{2}.$ Given that this ratio is $\frac{x}{2}$, we get $\frac{x}{2}=\frac{1}{2} \Rightarrow x=1.$

9. Final answer: $\boxed{1}$

The derived answer matches the stored correct answer.