The equation of the plane containing the straight line $\frac{x}{2}=\frac{y}{3}=\frac{z}{4}$

and perpendicular to the plane containing the straight lines $\frac{x}{3}=\frac{y}{4}=\frac{z}{2}$ and $\frac{x}{4}=\frac{y}{2}=\frac{z}{3}$ is

Vector along the normal to the plane containing the lines $\frac{x}{3}=\frac{y}{4}=\frac{z}{2}$ and $\frac{x}{4}=\frac{y}{2}=\frac{z}{3}$ is $8\widehat{i}-\widehat{j}-10\widehat{k}$

Vector perpendicular to the vectors $2\widehat{i}+3\widehat{j}+4\widehat{k}$ and $8\widehat{i}-\widehat{j}-10$ is $26\widehat{i}-52\widehat{j}+26\widehat{k}$

$\therefore$ Equation of required plane is

$26x-52y+26z=0\Rightarrow x-2y+z=0$