If lines x=y=z and x=y/2=z/3 and third line passing through (1,1,1) form a triangle of area  6 units, then point of intersection of third line with second line will be

x = y = z--- (1),    x=y2=z3 --- (2)Clearly point of intersection of (1) and (2) is (0,0,0)  D.r’s of (1) are (1, 1, 1)D.r’s of (2) are (1, 2, 3)Let θ be the angle between (1) and (2)cosθ=642,sinθ=642Let any point on second line be (λ,2λ,3λ)Third line passing through (1, 1, 1)(1, 1, 1) lies on (1)A = (1, 1, 1)Area of ΔOAB=12(OA)OBsinθ=123λ14×642=6⇒λ=2So B is (2,4,6)