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

$L: \dfrac{x}{2} = \dfrac{y}{3} = \dfrac{z}{4}$

$L_1: \dfrac{x}{3} = \dfrac{y}{4} = \dfrac{z}{2}, \quad L_2: \dfrac{x}{4} = \dfrac{y}{2} = \dfrac{z}{3}$

$\text{To find: plane } P \supset L, \quad P \perp \text{plane containing } L_1,\ L_2$

$\dfrac{x}{a} = \dfrac{y}{b} = \dfrac{z}{c} \Rightarrow \text{through origin}, \quad \vec{d} = (a,b,c)$

$\vec{d} = (2,3,4), \quad \vec{d}_1 = (3,4,2), \quad \vec{d}_2 = (4,2,3)$

$Q \supset L_1,\ L_2 \Rightarrow \vec{n}_Q = \vec{d}_1 \times \vec{d}_2$

n⃗_Q = î ĵ k̂ 3 4 2 4 2 3

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

$\vec{n}_Q = 8\hat{i} - \hat{j} - 10\hat{k}$

$\text{Let } P \text{ have normal } \vec{N}$

$P \supset L \Rightarrow \vec{N}\cdot\vec{d} = 0, \quad \vec{d} = (2,3,4)$

$P \perp Q \Rightarrow \vec{N}\cdot\vec{n}_Q = 0$

$\vec{N} \perp \vec{d} \ \text{and} \ \vec{N} \perp \vec{n}_Q \Rightarrow \vec{N} = \vec{d} \times \vec{n}_Q$

N⃗ = î ĵ k̂ 2 3 4 8 -1 -10

$\vec{N} = \hat{i}(-30+4) - \hat{j}(-20-32) + \hat{k}(-2-24)$

$\vec{N} = -26\hat{i} + 52\hat{j} - 26\hat{k}$

$\vec{N} = -26(\hat{i} - 2\hat{j} + \hat{k}) \Rightarrow \vec{N} \propto (1,-2,1)$

$L \text{ through origin} \Rightarrow P \text{ also through origin}$

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

$(1)\ (1,2,-2), \quad (2)\ (3,2,-2), \quad (3)\ (1,-2,1), \quad (4)\ (5,2,-4)$

$\text{Only } (3) \text{ has normal} \propto (1,-2,1)\ \checkmark$

$\vec{N}\cdot\vec{d} = 1(2) - 2(3) + 1(4) = 0 \ \checkmark$

$\vec{N}\cdot\vec{n}_Q = 1(8) - 2(-1) + 1(-10) = 0 \ \checkmark$

$\boxed{x - 2y + z = 0 \quad \Rightarrow \quad \text{option (3)}}$