The equation of a plane passing through the line of intersection of the planes x+2 y+3 z=2 , x-y+z=3 and at a distance 2 3 from the point (3,1,-1) is

Step-by-step solution

The equation of a plane passing through the intersection of the planes P₁: x + 2y + 3z - 2 = 0 and P₂: x - y + z - 3 = 0 is given by P₁ + P₂ = 0 . (x + 2y + 3z - 2) + (x - y + z - 3) = 0 (1 + )x + (2 - )y + (3 + )z - (2 + 3 ) = 0 The distance of this plane from the point (3, 1, -1) is given as 2 3 . Using the distance formula |Ax₀ + By₀ + Cz₀ + D| A^2 + B^2 + C^2 = d , we have: |(1 + )(3) + (2 - )(1) + (3 + )(-1) - (2 + 3 )| (1 + )^2 + (2 - )^2 + (3 + )^2 = 2 3 Simplifying the numerator: |3 + 3 + 2 - - 3 - - 2 - 3