The equation of the plane containing the line x+1 2 = y+2 1 = z-2 3 and the point (1,-1,3) is

Step-by-step solution

The plane contains the line x+1 2 = y+2 1 = z-2 3 and point (1, -1, 3) . Extracting line parameters: The symmetric form x-x₀ a = y-y₀ b = z-z₀ c gives point P₁ = (-1, -2, 2) with direction vector d = 2, 1, 3 . Forming additional vector: From P₁ to the given point P₂ = (1, -1, 3) , we obtain P₁P₂ = 2, 1, 1 . Normal vector calculation: The cross product d P₁P₂ yields: n = vmatrix i & j & k 2 & 1 & 3 2 & 1 & 1 vmatrix = -2, 4, 0 Simplifying gives n' = 1, -2, 0 . Plane equation: Using point P₂ and normal vector n' : 1(