The equation of the plane passing through the lines x−41=y−31=z−22 and x−31=y−2−4=z5

Here, the required plane is a (x - 4) + b (y- 3) + c (z -2) = 0

Also a+b+2c -0 and a-4b +5c-0 Solving, we have

$\begin{array}{l} \frac{a}{5 + 8} = \frac{b}{2 - 5} = \frac{c}{- 4 - 1} = k \\ \frac{a}{13} = \frac{b}{- 3} = \frac{c}{- 5} = k \end{array}$

Therefore, the required equation of plane is -13x+3y+5z+33=0

The equation of the plane passing through the lines $\frac{x - 4}{1} = \frac{y - 3}{1} = \frac{z - 2}{2}$ and $\frac{x - 3}{1} = \frac{y - 2}{- 4} = \frac{z}{5}$