Two vectors $a \hat{i}+b \hat{j}+\hat{k}$ and $2 \hat{i}-3 \hat{j}+4 \hat{k}$ are perpendicular to each…

Given vectors $\mathrm{V_1} = a\hat{i} + b\hat{j} + \hat{k}$ and $\mathrm{V_2} = 2\hat{i} - 3\hat{j} + 4\hat{k}$ are perpendicular, so their dot product equals zero: $2a - 3b + 4 = 0$
This simplifies to $2a - 3b = -4$

The second condition gives $3a + 2b = 7$

Solving this system by elimination, multiply the first equation by 2 and the second by 3:

$4a - 6b = -8$
$9a + 6b = 21$

Adding eliminates $b$ and yields $13a = 13$, so $a = 1$

Substituting into $3a + 2b = 7$ gives $3 + 2b = 7$, hence $b = 2$

The ratio $\frac{a}{b} = \frac{1}{2}$ equals $\frac{x}{2}$, so $x = 1$

Answer: $\boxed{C}$