If the equation of plane containing lines $\frac{x-1}{2}=\frac{y+1}{-1}=\frac{z-1}{3}$and  $\frac{x-3}{-4}=\frac{z-2}{2}=\frac{z-1}{-6}$is  $9x+by+cz=d$ then $\left(b+c+d\right)$  is

The equation of the plane containing two parallel lines  $\frac{x-x_1}{a_1}=\frac{y-y_1}{b_1}=\frac{z-z_1}{c_1}$and  $\frac{x-x_2}{a_1}=\frac{y-y_2}{b_1}=\frac{z-z_2}{c_1}$is $\left|\begin{array}{ccc}x-x_1& y-y_1& z-z_1\\ x_2-x_1& y_2-y_1& z_2-z_1\\ a_1& b_1& c_1\end{array}\right|=0$

The given two lines are parallel, because their direction ratios are proportional with$〈2,-1,3〉$

Hence, the equation of the plane containing the both lines is $\left|\begin{array}{ccc}x-1& y+1& z-1\\ 3-1& 2+1& 1-1\\ 2& -1& 3\end{array}\right|=0$

Simplifying 

 $\begin{array}{c}\left(x-1\right)\left(9\right)-\left(y+1\right)\left(6\right)+\left(z-1\right)\left(-2-6\right)=0\\ 9x-9-6y-6-8z+8=0\\ 9x-6y-8z-7=0\\ 9x-6y-8z=7\end{array}$      

Compare this equation with  $ax+by+cz=d$

Hence,$b+c+d=-6-8+7=\boxed{-7}$