Let the line x/3 = y/1 = z+2/2 lie in the plane x + 3y - αz + β = 0, then (α, β) equals (7,-5), (-6,7), (2,-56), (-6,6)

If the given line lie in the plane x + 3y - αz + β = 0
2 + 3 × 1 - α( -6) + β = 0
2α + β = -5 (i)
Drs of normal = 1, 3, -α
Drs of line = 3, 1, 2
line is perpendicular to normal
=> 3(1) + 1(3) - α(2) = 0
=> 2α = 6
=> α = 3
From (i)
2(3) + β = -5
β = -5 - 6 = -11
Then (α, β) = (3, -11)
Let's check the options
If (α, β) = (7, -5)
2(7) + (-5) = 9 ≠ -5
If (α, β) = (-6, 7)
2(-6) + 7 = -5
3(1) + 1(3) - (-6)(2) = 3 + 3 + 12 = 18 ≠ 0
If (α, β) = (-6, 6)
2(-6) + 6 = -6 ≠ -5
If (α, β) = (2, -56)
2(2) + (-56) = -52 ≠ -5
The direction ratios of the line are 3, 1, 2.
The direction ratios of the normal to the plane are 1, 3, -α.
Since the line lies in the plane, the line is perpendicular to the normal to the plane.
Therefore, the dot product of the direction ratios of the line and the normal to the plane is 0.
3(1) + 1(3) + 2(-α) = 0
3 + 3 - 2α = 0
6 - 2α = 0
2α = 6
α = 3
Substituting x = 3, y = 1, z = -6 in the equation of the plane:
3 + 3(1) - 3(-2) + β = 0
3 + 3 + 6 + β = 0
12 + β = 0
β = -12
Let's check the given options:
2 + 3(1) - α(-2) + β = 0
5 + 2α + β = 0
If α = 3, then 5 + 6 + β = 0, which gives β = -11.
If α = -6, then 5 - 12 + β = 0, which gives β = 7.
If α = 2, then 5 + 4 + β = 0, which gives β = -9.
If α = -6, then 5 - 12 + β = 0, which gives β = 7.
Hence (α, β) = (-6, 7)