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

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

Given that the plane is $\perp$ to the line joining $A(-3,1,2)\:\:and\:\:B(2,3,4)$

$\therefore\: \overline {AB}=(5,2,2)$ becomes normal to the plane.

$\Rightarrow\:(a,b,c)=(5,2,2)$

$\therefore$ The equation of the plane becomes $ 5x+2y+2z+d=0$

But given that the plane passes through $(-1,2,1)$

$\therefore\:-5+4+2+d=0$

$\Rightarrow\:d=-1$

$\Rightarrow\:$ The equation of the plane is $5x+2y+2z-1=0$