For planes 2x+y+z+1=0‌‌‌ and ‌‌‌2x+y+z+α=0, the distance between them is ‌ Distance ‌=‌
|α−1|
√22+12+12
=‌
|α−1|
√6
. Given this distance is 3 units: ‌
|α−1|
√6
=3⇒|α−1|=3√6. So, α−1=±3√6⇒α=1+3√6‌‌‌ or ‌‌‌α=1−3√6. Product of all possible values of α : (1+3√6)(1−3√6)=1−(3√6)2=1−9⋅6=1−54=−53. -53