Do you mean findanotherpoint, say (x, y, z), on that same surface so that the tangent plane at (x, y, z) is perpendicular to the tangent plane at (6, 3, 2)?

Okay, as you say. At (6, 3,2 ) that is <2(6), 4(3), 6(2)= <12, 12, 12> which is, as I think you mean, parallel to <1, 1, 1>. And that is perpendicular to the tangent plane at (6, 3, 2). Two planes are perpendicular if and only if their normal vectors are perpendicular. You want to find (x, y, z) such that and, of course, such that .