Concept:For a line to lie in a plane, its direction vector must be perpendicular to the plane's normal vector, and any point on the line must satisfy the plane's equation.Explanation:The line passes through the point (2,1,−2).Its direction vector is (3,−5,2).The normal vector of the plane x+3y−αz+β=0 is (1,3,−α).Since the point (2,1,−2) lies on the plane,2+3(1)−α(−2)+β=0⇒5+2α+β=0⇒2α+β=−5 ... (i)For the line to lie in the plane, its direction vector must be perpendicular to the normal vector:(3)(1)+(−5)(3)+(2)(−α)=0⇒3−15−2α=0⇒−12−2α=0⇒α=−6Substitute α=−6 in (i):2(−6)+β=−5⇒−12+β=−5⇒β=7Thus, (α,β)=(−6,7).Answer:(α,β)=(−6,7)Correct option: B. (−6,7)