Let the equation of the plane be $ax+by+cz+d=0$

$\Rightarrow\:$ D.R. of the normal to the plane is $ (a,b,c)$

Given: This plane is $\perp$ to the plane $3x-y+2z-3=0$

$\therefore\: (a,b,c).(3,-1.2)=0$ $ \Rightarrow\:3a-b+2c=0.....(i)$

Also given that the plane is parallel to the line $\large\frac{x-1}{3}=\frac{y}{2}=\frac{z}{-2}=\lambda\:\: (say)$

$\Rightarrow\:$ Normal to the plane is $\perp$ to the line.

$\Rightarrow\:(a,b,c).(3,2,-2)=0\:\:\Rightarrow\:3a+2b-2c=0.......(ii)$

It is given that the plane passes through the point $(1,2,3)$.

Solving $(i)\:\:and\:\:(ii)$ we get

$\large\frac{a}{2-4}=\frac{-b}{-6-6}=\frac{c}{6+3}$

$\Rightarrow\:a=-2,\:b=12, \:c=9$

Substituting the values in the equation of the plane we get

$-2x+12y+9z+d=0$

But given that the point $(1,2,3)$ lies on the plane.

$\therefore\:-2+24+27+d=0$

$\Rightarrow\:d=-49$

$\therefore$ The required equation of the plane is

$2x-12y-9z+49=0$

In dot product form the equation is : $\overrightarrow r.(2\hat i-12\hat j-9\hat k)+49=0$