Plane 1 : $$ax + by + cz = 0$$ contains line $${x \over 2} = {y \over 3} = {z \over 4}$$. Therefore,

$$2a + 3b + 4c = 0$$ ..... (1)

Plane 2 : $$a'x + b'y + c'z = 0$$ is perpendicular to plane containing line $${x \over 3} = {y \over 4} = {z \over 2}$$ and $${x \over 4} = {y \over 2} = {z \over 3}$$.

Hence, $$3a' + 4b' + 2c' = 0$$ and $$4a' + 2b' + 3c' = 0$$.

$$ \Rightarrow {{a'} \over {12 - 4}} = {{b'} \over {8 - 9}} = {{c'} \over {6 - 16}}$$

$$ \Rightarrow 8a - b - 10c = 0$$ .... (2)

From Eq. (1) and (2), we get

$${a \over { - 30 + 4}} = {b \over {32 + 20}} = {c \over { - 2 - 24}}$$

$$\Rightarrow$$ Equation of plane is $$x - 2y + z = 0$$