Equation of the plane containing the lines $\frac{x}{1}=\frac{y-2}{3}=\frac{z+4}{-1}$ and $\frac{x-4}{2}=\frac{y}{3}=\frac{z}{1}$ is, :

None of these

Equation of any plane containing the line $\frac{x}{1}=\frac{y-2}{3}=\frac{z+4}{-1}$ is ax + b(y − 2) + c(z + 4) = 0

where a + 3b – c = 0

This plane will also contain the second line if

2a – 3b + c = 0

and 4ab(0 – 2) + c(0 + 4) = 0

Solving these equation, we get

a = 0, b = 0, c = 0

That means the given lines are non–coplanar.