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

$L_1:\ \dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4} \;\Rightarrow\; \vec d_1=(2,3,4)$

$L_2:\ \dfrac{x}{3}=\dfrac{y}{4}=\dfrac{z}{2} \;\Rightarrow\; \vec d_2=(3,4,2)$

$L_3:\ \dfrac{x}{4}=\dfrac{y}{2}=\dfrac{z}{3} \;\Rightarrow\; \vec d_3=(4,2,3)$

$\text{Plane } P \supset L_1, \quad P \perp \text{plane}(L_2, L_3)$

$\text{To find: equation of } P$

$1.\ \vec n_2 = \vec d_2 \times \vec d_3$

$2.\ \vec n = (a,b,c):\ \vec n\cdot\vec d_1 = 0,\ \vec n\cdot\vec n_2 = 0$

$3.\ \text{solve for } \vec n,\ \text{write the plane}$

n⃗₂ = d⃗₂ × d⃗₃ = î ĵ k̂ 3 4 2 4 2 3

$= \hat{i}(12-4) - \hat{j}(9-8) + \hat{k}(6-16)$

$\vec n_2 = (8,\ -1,\ -10)$

$\text{Plane}(L_2,L_3):\ 8x - y - 10z = 0$

$P:\ ax+by+cz = 0 \quad (\text{contains } L_1 \text{ through } O)$

$\vec n\cdot\vec d_1 = 0:\quad 2a+3b+4c=0$

$\vec n\cdot\vec n_2 = 0:\quad 8a - b - 10c = 0$

$b = 8a - 10c$

$2a + 3(8a-10c) + 4c = 0 \;\Rightarrow\; 26a - 26c = 0$

$a = c, \quad b = -2a$

$\vec n = (a,\,-2a,\,a) \parallel (1,\,-2,\,1)$

$P:\ x - 2y + z = 0$

$\vec n\cdot\vec d_1 = 1(2)+(-2)(3)+1(4) = 0 \ \checkmark$

$\vec n\cdot\vec n_2 = 1(8)+(-2)(-1)+1(-10) = 0 \ \checkmark$

$\boxed{x - 2y + z = 0}$

$\text{Answer: option (3)}$