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

The DR's of normal to the plane containing \(\frac{x}{3}=\frac{y}{4}=\frac{z}{2}\) and \(\frac{x}{4}=\frac{y}{2}=\frac{z}{3}\). \(\Rightarrow \quad \overrightarrow{\mathbf{n}}_1=\left|\begin{array}{ccc}\hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ 3 & 4 & 2 \\ 4 & 2 & 3\end{array}\right|=(8 \hat{\mathbf{i}}-\hat{\mathbf{j}}-10 \hat{\mathbf{k}})\)
Also, equation of plane containing \(\frac{x}{2}=\frac{y}{3}=\frac{z}{4}\) and DR's of normal to be \(\overrightarrow{\mathbf{n}}_2=a \hat{\mathbf{i}}+b \hat{\mathbf{j}}+c \hat{\mathbf{k}}\)



\[
\begin{array}{ll}
\Rightarrow & a x+b y+c z=0 \\
\text { where, } \overrightarrow{\mathbf{n}}_1 \cdot \overrightarrow{\mathbf{n}}_2=0 \\
\Rightarrow & 8 a-b-10 c=0 \\
\text { and } & \overrightarrow{\mathbf{n}}_2 \perp(2 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}+4 \hat{\mathbf{k}}) \\
\Rightarrow & 2 a+3 b+4 c=0
\end{array}
\]
From Eqs. (ii) and (iii), we get
\[
\begin{aligned}
& \frac{a}{-1 \quad-10}=\frac{b}{8}=\frac{c}{-1} \\
\Rightarrow \quad \frac{a}{-4+30} & =\frac{b}{-20-32}=\frac{c}{24+2} \\
\Rightarrow \quad \quad \quad \frac{a}{26} & =\frac{b}{-52}=\frac{c}{26} \\
\Rightarrow \quad \quad \quad \quad \frac{a}{1} & =\frac{b}{-2}=\frac{c}{1}
\end{aligned}
\]
From Eqs. (i) and (iv), required equation of plane, is \(x-2 y+z=0\).