The equation of the plane containing the two lines $\frac{x-1}{2}=\frac{y+1}{-1}=\frac{z}{3}\text{ and }\frac{x}{2}=\frac{y-2}{-1}=\frac{z+1}{-3}$ is:

Any plane through the first line may be written as $a(x-1)+b(y+1)+c\left(z\right)=0 ...\left(i\right)$

Where,  $2a-b+3c=0 ...\left(ii\right)$

It will pass through the second line, if the point $(0, 2, -1)$on the second line also lies on (i) 

i.e. if $a(0-1)+b(2+1)+c(-1)=0,$

i.e., $-a+3b-c=0 ...\left(iii\right)$

Solving (ii) and (iii), we get $\frac{a}{-8}=\frac{b}{-1}=\frac{c}{5}$

i.e $\frac{a}{8}=\frac{b}{1}=\frac{c}{-5}$

$\therefore$ Required plane is  $8(x-1)+1(y+1)-5\left(z\right)=0$

$\Rightarrow 8x+y-5z-7=0.$