Equation of the plane containing the straight line $\frac{x}{3}=\frac{y}{2}=\frac{z}{4}$ and perpendicular…
Equation of the plane containing the straight line $\frac{x}{3}=\frac{y}{2}=\frac{z}{4}$ and perpendicular to the plane containing the straight lines $\frac{x}{4}=\frac{y}{3}=\frac{z}{2}$ and $\frac{\dot{x}}{2}=\frac{y}{-4}=\frac{z}{3}$ is
$\quad 6 x-67 y-29 \mathrm{z}=0$
$6 x+67 y-29 z=0$
$\quad 6 x-67 y+29 z=0$
$6 x+67 y+29 z=0$
Solution
Equation of the plane containing $\frac{x}{4}=\frac{y}{3}=\frac{z}{2}$ and $\frac{x}{2}=\frac{y}{-4}=\frac{z}{3}$ is
$\begin{aligned}
& \left|\begin{array}{ccc}
x & y & z \\
4 & 3 & 2 \\
2 & -4 & 3
\end{array}\right|=0 \\
& \Rightarrow 17 x-8 y-22 z=0
\end{aligned}$ Now required plane is perpendicular to this plane.
Consider option (C)
$(17)(6)+(-8)(-67)+(-22)(29)=0$
$\therefore \quad$ Option (C) is correct.