The equation of a plane passing through (1, 2, -3), (0, 0, 0) and perpendicular to the plane 3x – 5y + 2z = 11, is :

$3 x+y+\frac{5}{3} z=0$ 4x + y + 2z = 0 $3 x-y+\frac{z}{3}$ x + y + z = 0

x + y + z = 0

Let the required plane be Ax + by + cz = 0 We have 3a – 5b + 2c = 0, a + 2b – 3c = 0 $\Rightarrow \frac{a}{15-4}=\frac{b}{2+9}=\frac{c}{6+5}$ $\Rightarrow a: b: c=11: 11: 11$ Thus plan is x + y + z = 0