Answer is (D) 3/x + 2/y + 1/z = 6
Equation of plane is given as
x/a + y/b + z/c = 1
Therefore, the equation of plane passing through (3, 2, 1) is
3/a + 2/b + 1/c = 1
Points on axes are A(a, 0, 0), B(0, b, 0) and C(0, 0, c).
Therefore, the locus of intersection point of plane through point A, B and C. Parallel to yz-, zx- and xy- plane, respectively, is
3/x + 2/y + 1/c = 1