The correct option is B −20
3x2−5xy+2y2=0
t=yx,
2t2−5t+3=0
t=5±√25−244
t=5±14
t=32,1
given 3x2−5xy+2y2=0 and 6x2−xy+ky2=0 have one line in common
So, If t=32,
kt2−1t+6=0
k94−32+6=0
k=−2 ----(1)
If, t=1, k−1+6=0
k=−5 ----(2)
When k=−2, k2+7k−10=4−14−10
=−20
When, k=−5 k2+7k−10=25−35−10
k2+7k−10=−20