1.Consider three lines L1:5x−y+4=0L_1: 5x - y + 4 = 0L1:5x−y+4=0, L2:3x−y+5=0L_2: 3x - y + 5 = 0L2:3x−y+5=0 and L3:x+y+8=0L_3: x + y + 8 = 0L3:x+y+8=0. If these lines form a triangle ABC, then the sum of the squares of the tangent of all the interior angles of the triangle isa.40164\frac{401}{64}64401b.39964\frac{399}{64}64399c.11164\frac{111}{64}64111d.11364\frac{113}{64}64113Login to continueOnly logged in users canattempt or see the solution.