A point on the plane passes through the points (1,-1,6), (0,0,7) and perpendicular to the plane $x-2y+z=6$ is

plane equation $a\left(x-x_1\right)+b\left(y-y_1\right)+c\left(z-z_1\right)=0$

Passing $\left(1,-1,6\right)\left(0,0,7\right)$

i.e $a\left(0-1\right)+b\left(0+1\right)+c\left(7-6\right)=0$

$-a+b+c=0---\left(1\right)$

And plane $\perp erto$ $x-2y+z=6$ is

$a-2b+c=0---------\left(2\right)$

solve (1) and (2)

$\begin{array}{cccc} & a& b& c\\ 1& 1& -1& 1\\ -2& 1& 1& -2\end{array}$

$\frac{a}{1+2}=\frac{b}{1+1}=\frac{c}{2-1}$

$\left(a,b,c\right)=\left(3,2,1\right)$

$\therefore$plane equation 

$3\left(x-1\right)+2\left(y+1\right)+1\left(z-6\right)=0$

$3x+2y+z-7=0$

Option verification

Option (2) satisfies

i.e (1,1,2)