x+y+z=6 .......(i) x+2y+3z=10 .....(ii) 3x+2y+λz=µ ......(iii) From (i) and (ii), If z=0⇒x+y=6 and x+2y=10 ⇒y=4,x=2 (2,4,0) If y=0⇒x+z=6 and x+3z=10 ⇒z=2 and x=4 (4,0,2) So, 3x+2y+λz=µ, must pass through (2,4,0) and (4,0,2) So, 6+8=µ⇒µ=14 and 12+2λ=µ 12+2λ=14⇒λ=1 So, µ−λ2=14−1=13