If for some α and β in R , the intersection of the following three planes x + 4 y - 2 z = 1 x + 7 y - 5 z = β x + 5 y + α z = 5 is a line in R 3 , then α + β is equal to:

Step-by-step solution

∆ = 0 ⇒ 1 4 - 2 1 7 - 5 1 5 α = 0 7 α + 25 - 4 α + 10 + - 20 + 14 = 0 3 α + 9 = 0 ⇒ α = - 3 Also D z = 0 ⇒ 1 4 1 1 7 β 1 5 5 = 0 1 35 - 5 β - 15 + 1 4 β - 7 = 0 β = 13