Let  $L_1:\frac{x-6}{3}=\frac{y-7}{-1}=\frac{z-4}{1}and\frac{x}{-3}=\frac{y+9}{2}=\frac{z-2}{4}.$  If  $L_{1}and{L}_2$ are skew lines such that P and Q are two points nearest to each other lying on lines $L_{1}and{L}_2$ respectively. If image of P with respect to plane $3x+3y-z=11 is R(a,b,c) then a+b+\frac{c}{2}$  is

Direction ratio of  $\text{PQ}$
 $(-3r_2-3r_1-6,2r_2+r_1+6,4r_2-r_1-2)$
Now,  $-3(-3r_2-3r_1-6)+2(2r_2+r_1-16)+(4r_2-r_1-2)=0$
 $$\begin{gathered} \therefore 7r_1+29r_2=22 ---\left(1\right) \\ 3(-3r_2-3r_1-6)-(2r_2+r_1-16)+(4r_2-r_1-2)=0 \\ 7r_2+11r_1=-4 ---\left(2\right) \therefore r_1=-1,r_2=1 \\ a=-3, b=2,c=5 \\ -a+b+\frac{c}{2}=3+2+25=7.5 \end{gathered}$$