Plane containing the straight line

 $\frac{\mathrm{x}}{2}=\frac{\mathrm{y}}{3}=\frac{\mathrm{z}}{4}$  is   Ax + By + Cz = 0             ....(1)

The vector $\overrightarrow{\mathrm{V}}=\mathrm{A}\hat{\mathrm{i}}+\mathrm{B}\hat{\mathrm{j}}+\mathrm{C}\hat{\mathrm{k}}$  is perpendicular to $2\hat{\mathrm{i}}+3\hat{\mathrm{j}}+4\hat{\mathrm{k}}$

Also (1) is perpendicular to the plane Q containing the lines

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

hence vector normal to Q = $\left|\begin{array}{ccc}\hat{\mathrm{i}}& \hat{\mathrm{j}}& \hat{\mathrm{k}}\\ 3& 4& 2\\ 4& 2& 3\end{array}\right|=\hat{\mathrm{i}}\left(8\right)-\hat{\mathrm{j}}\left(1\right)+\hat{\mathrm{k}}(-10)=8\hat{\mathrm{i}}-\hat{\mathrm{j}}-10\hat{\mathrm{k}}$

Now $\overrightarrow{\mathrm{V}}$  is perpendicular to  $2\hat{\mathrm{i}}+3\hat{\mathrm{j}}+4\hat{\mathrm{k}}$  as well as  $8\hat{\mathrm{i}}-\hat{\mathrm{j}}-10\hat{\mathrm{k}}\Rightarrow \overrightarrow{\mathrm{V}}$ is collinear with

 $\left|\begin{array}{ccc}\hat{\mathrm{i}}& \hat{\mathrm{j}}& \hat{\mathrm{k}}\\ 2& 3& 4\\ 8& -1& -10\end{array}\right|=-26\hat{\mathrm{i}}+52\hat{\mathrm{j}}-26\hat{\mathrm{k}}$

 $\therefore$    A, B, C are  $\propto$ to 1, – 2, 1

Hence required plane is   x – 2y + 2 = 0 Ans.