The correct answer is (d) a₁x + b₁y + c₁z + d₁ = a₂x + b₂y + c₂z + d₂ + (a₁ - a₂)x + (b₁ - b₂)y + (c₁ - c₂)z.
How to find the equation of the plane
To find the equation of the plane through the intersection of P₁ and P₂ and the point (3, 2, 1), find the direction vector of the intersection line of P₁ and P₂ and then use that along with the given point to construct the equation of the plane.
The direction vector of the intersection line of P₁ and P₂ can be found by taking the cross product of the normal vectors of the two planes.
Let's denote the normal vectors of P₁ and P₂ as n₁ and n₂, respectively.
n₁ = (a₁, b₁, c₁)
n₂ = (a₂, b₂, c₂)
The direction vector of the intersection line, d, is given by:
d = n₁ × n₂
Now, construct the equation of the plane using the point (3, 2, 1) and the direction vector d.
The equation of the plane passing through the point (3, 2, 1) and with direction vector d can be written as:
(a₁ - a₂)(x - 3) + (b₁ - b₂)(y - 2) + (c₁ - c₂)(z - 1) = 0
Expand this equation
(a₁ - a₂)x + (b₁ - b₂)y + (c₁ - c₂)z + (3(a₂ - a₁) + 2(b₂ - b₁) + (c₂ - c₁)) = 0
Simplify the equation further
(a₁ - a₂)x + (b₁ - b₂)y + (c₁ - c₂)z + (a₁x + b₁y + c₁z + d₁) - (a₂x + b₂y + c₂z + d₂) = 0
Combine like terms
(a₁ - a₂)x + (b₁ - b₂)y + (c₁ - c₂)z + (d₁ - d₂) = 0
Therefore, the correct answer is (d) a₁x + b₁y + c₁z + d₁ = a₂x + b₂y + c₂z + d₂ + (a₁ - a₂)x + (b₁ - b₂)y + (c₁ - c₂)z.