A line through  A(−5,−4) with slope tanθ  meets the lines x+3y+2=0, 2x+y+4=0, x−y−5=0 at B, C, D respectively, such that  15AB2+10AC2=6AD2then

A line through  A(−5,−4) with slope tanθ is x+5cosθ=y+4sinθ=r Any point on the line is =(−5+rcosθ,−4+rsinθ)If this lies on x+3y+2=0, we have −5+rcosθ+3(−4+rsinθ)+2=0∴r=15AB=cosθ+3sinθsimilarly, we get, 10AC=2cosθ+sinθ and 6AD=cosθ−sinθFrom conditions, (cosθ+3sinθ)2+(2cosθ+sinθ)2=(cosθ−sinθ)2⇒(2cosθ+3sinθ)2=0⇒tanθ=−23