Equation of plane passing through a point (x1,y1,z1) is a(x−x1)+b(y−y1)+c(z−z1)=0 Here, (x1,y1,z1)=(1,2,3) So,a(x−1)+b(y−2)+c(z−3)=0 Now, Y-axis lies on it. Direction ratio of Y-axis is (0,1,0). Normal vector to the plane =a
∧
i
+b
∧
j
+c
∧
k
So, the normal vector of the plane will be perpendicular to direction ratio of Y-axis. a⋅0+b⋅1+c⋅0=0⇒b=0 Equation of plane becomes a(x−1)+c(z−3)=0 Now, x=0,z=0 also satisfies the equation. a(0−1)+c(0−3)=0 ⇒t−a−3c=0⇒a=−3c So, −3c(x−1)+c(z−3)=0 −3x+3+z−3=0 [as,C≠0] ⇒3x−z=0