The plane containing the lines $\frac{x+1}{3}=\frac{y+3}{5}=\frac{z+5}{7}$  and $\frac{x-2}{1}=\frac{y-4}{4}=\frac{z-6}{7}$ passes through 

Let the equation of the plane be $lx + my + mz + d$
$= 0, 3l+5m+7n=0$ and $l+4m+7n=0$
$\Rightarrow l/1=m/-2=n/1$ and the equation of the plane is $x - 2y + z + d = 0$ since it passes through $(- 1, - 3, - 5)$ and $(2, 4, 6), d = 0$ and the plane passes through (0, 0, 0).