Equation Of A Plane Through Three Points

The relevant idea is the intercept form of a plane, x/a + y/b + z/c = 1, which directly applies when a plane cuts the axes at known intercepts; equivalently one can use the determinant form through three points. This is a classic JEE Advanced shortcut. The given points are precisely the unit intercepts on the x, y and z axes, so a = b = c = 1, yielding x/1 + y/1 + z/1 = 1, that is x + y + z = 1. To verify by the normal approach, vectors in the plane are (0,1,0)-(1,0,0) = (-1, 1, 0) and (0,0,1)-(1,0,0) = (-1, 0, 1); their cross product is (1, 1, 1), a valid normal, and using point (1,0,0) gives 1(x-1)+1y+1z = 0, i.e. x + y + z = 1. Option x + y + z = 0 fails the points. Option x + y + z = 3 misplaces the constant. Option x - y + z = 1 uses a wrong normal. This rests on the intercept-form plane theorem. Plausibility check: each of the three given points individually satisfies x + y + z = 1, confirming the plane indeed contains them all.