If the plane x 2 + y 3 + z 6 =1 cuts the co - ordinate axes at points A, B, C respectively, then area of the triangle ABC is

Step-by-step solution

The plane given by x 2 + y 3 + z 6 = 1 intercepts the coordinate axes at A = (2, 0, 0) , B = (0, 3, 0) , and C = (0, 0, 6) , forming triangle ABC . Vectors representing two sides are AB = (-2, 3, 0) and AC = (-2, 0, 6) . The cross product yields AB AC = (18, 12, 6) with magnitude | AB AC | = 18^2 + 12^2 + 6^2 = 504 = 6 14 . The area of triangle ABC is 1 2 6 14 = 3 14 square units. Alternatively, using the intercept formula 1 2 (ab)^2 + (bc)^2 + (ca)^2 with a=2 , b=3 , c=6 gives the same result: 1 2 (6)^2 + (18)^2 +