Three Dimensional Geometry Part 2

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