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 $\frac{x}{3}=\frac{y}{4}=\frac{z}{2}\text{ and }\frac{x}{4}=\frac{y}{2}=\frac{z}{3}$ is

$\text{ Plane }1:ax+by+cz=0\text{ contains line }\frac{x}{2}=\frac{y}{3}=\frac{z}{4}$

$\therefore 2a+3b+4c=0----\left(1\right)$

$\text{ Plane 2: }{a}^{\mathrm{\prime }}x+b^{\mathrm{\prime }}y+c^{\mathrm{\prime }}z=0\text{ is perpendicular to plane containing }$

$\text{ lines }\frac{x}{3}=\frac{y}{4}=\frac{z}{2}\text{ and }\frac{x}{4}=\frac{y}{2}=\frac{z}{3}$

$\begin{array}{l}\therefore 3a^{\mathrm{\prime }}+4b^{\mathrm{\prime }}+2c^{\mathrm{\prime }}=0\text{ and }4a^{\mathrm{\prime }}+2b^{\mathrm{\prime }}+3c^{\mathrm{\prime }}=0\\ \Rightarrow \frac{a^{\mathrm{\prime }}}{12-4}=\frac{b^{\mathrm{\prime }}}{8-9}=\frac{c^{\mathrm{\prime }}}{6-16}\\ \Rightarrow 8a-b-10c=0----\left(2\right)\end{array}$

From (i) and (ii), we get

$\begin{array}{l}\frac{a}{-30+4}=\frac{b}{32+20}=\frac{c}{-2-24}\\ \Rightarrow \text{ Equation of plane }x-2y+z=0\end{array}$