The value of m for which the line $\frac{x-1}{2}=\frac{y-1}{3}=\frac{z-1}{m}$ is perpendicular to  normal to the plane $\vec{r}.(2\hat{i} + 3\hat{j} + 4 \hat{k}) = 0, $ is

$-\frac{13}{4}$

Given line is parallel to the vector $\vec{b} = 2\hat{i} + 3\hat{j} + m\hat{k}$ and given plane is normal to $\vec{n} = 2\hat{i} + 3\hat{j} + 4\hat{k}$. If $\vec{b}$ and $\vec{n}$ are perpendicular , then 

$\vec{b}.\vec{n} = 0 ⇒ 4 + 9 + 4m = 0 ⇒ m = -\frac{13}{4}$