Right option is (d) 2x+y≤6, x≥0, y≥0
To explain I would say: Since region involves 1^st quadrant so x≥0, y≥0.
Two points on line are (0,6) and (3,0).
(y-6)/(0-6) = (x-0)/(3-0)
=>(y-6)/(-6) = x/3
=>y-6=-2x => 2x+y=6
2x+y≤6 since (0,0) should also satisfy.
So, 2x+y≤6, x≥0, y≥0.