Find the equation of the plane which passes through the points (0,1,2) and (-1,0,3) and is perpendicular to the plane 2 x+3 y+z=5 .

Step-by-step solution

Since, we know that the equation of plane passing through a point (x₁, y₁, z₁ ) is A (x-x₁ )+B (y-y₁ )+C (z-z₁ )=0 Equation of plane passing through (0,1,2) is A(x-0)+B(y-1)+C(z-2)=0 ...(i) Plane (i) passes through (-1,0,3) . Hence, A(-1-0)+B(0-1)+C(3-2)=0-A-B+C=0 A+B-C=0 ...(ii) If plane (i) is perpendicular to 2 x+3 y+z=5 2 A+3 B+C=0 ...(iii) [by using condition of perpendicularly] On solving Eqs. (ii) and (iii), A 1 1-3(-1) = B 2(-1)-1 1 = C 1 3-2 1 A 4 = B -3 = C 1 Using these proportional values of A, B, C in