Equation of a line is $\frac{x-3}{2}=\frac{4-y}{3}=\frac{2-z}{4}$ then its direction cosines are :

$\frac{2}{\sqrt{29}}, \frac{-3}{\sqrt{29}},\frac{-4}{\sqrt{29}}$

The correct answer is Option (4) → $\frac{2}{\sqrt{29}}, \frac{-3}{\sqrt{29}},\frac{-4}{\sqrt{29}}$

first converting eq. to std. form

$\frac{x-3}{2}=\frac{4-y}{3}=\frac{2-z}{4}$

DR's = 2, -3, -4

DC's = $\frac{2}{\sqrt{2^2+(-3)^2+(-4)^2}},\frac{-3}{\sqrt{2^2+(-3)^2+(-4)^2}},\frac{-4}{\sqrt{2^2+(-3)^2+(-4)^2}}$

DC's = $\frac{2}{\sqrt{29}}, \frac{-3}{\sqrt{29}},\frac{-4}{\sqrt{29}}$