If  $\frac{x-4}{1}=\frac{y-2}{1}=\frac{z-7}{2}$  lies in the plane  $ax+by+z=7$, then  $a+b=$

Given that, line  $\frac{x-4}{1}=\frac{y-2}{1}=\frac{z-7}{2}$
Line is passing through point $P(4,+2,7)$  and direction ratios $(1,1,2)$ and lies on the given plane  $ax+by+z=7$
So, point will satisfy the equation of plane
 $a\left(4\right)+b(+2)+\left(7\right)=7$
 $4a+2b=0 \mathrm{..................}\left(i\right)$
Now direction ratios of the line and direction ratios of the plane will be perpendicular so their product must be zero
 $a_1{a}_2+b_1{b}_2+c_1{c}_2=\left(1\right)\left(a\right)+\left(1\right)\left(b\right)+\left(2\right)\left(1\right)$
  $0=a+b+2$
or  $a+b=-2$