Find the equation of the plane passing through the intersection of line $$\frac{x-2}{3}=\frac{y+1}{4}=\frac{z-2}{2}$$ and the plane $$x-y+z=5$$
and parallel to a vector with direction ratios $<2,3,-2>$
Now point of intersection of given plane and given line is $(2,-1,2)$ and direction normal of required plane will be perpendicular to $<2,3,-2>$ but how would I get a unique equation of required plane?
So you want a plane through $\;(2,-1,2)\;$ and parallel to $\;(2,3,-2)\;$ , so for any $\;(a,b,c)\in\Bbb R^3\;$ the following fulfills the conditions:
$$(2,-1,2)+t(2,3,-2)+s(a,b,c)\;,\;\;t,s\in\Bbb R\;$$