Find the equation of the plane passing through the point and is normal to a plane

Question

The question is: Find the equation of the plane passing through the point (2, 1, 3) and is normal to the plane 3x - 7y + 5z = -55.

I tried using the formula r.n = a.n

Doing so, I get: (x,y,z).(3,-7,5) = (2,1,3).(3,-7,5)

Hence, the equation is, 3x - 7y + 5z = 14.

I'm not sure if this correct because the normal that I found for my equation is parallel to the plane but not perpendicular to it.

Answer 1

Let's denote by $\alpha:Ax+By+Cz=D$ a plane which solves the question.

The plane contains the point $P(2,1,3).$ So it is $2A+B+3C=D.$

Now, it is perpendicular to the plane $\pi:3x - 7y + 5z = -55.$ So, we have that $(A,B,C)\cdot (3,-7,5)=0.$ That is: $3A-7B+5C=0.$

We get $C=\frac{7B-3A}{5}$ and $D=2A+B+\frac{21B-9A}{5}.$

So the infinitely many solutions are given by

$$Ax+By+\frac{7B-3A}{5}z=2A+B+\frac{21B-9A}{5}$$ where $(A,B)\in \mathbb{R}^2, (A,B)\ne (0,0).$ In other words

$$5Ax+5By+(7B-3A)z=A+26B$$ where $(A,B)\in \mathbb{R}^2, (A,B)\ne (0,0).$

One particular solution ($A=B=1$) is $$5x+5y+4z=27.$$

Answer 2

Pick a vector on the plan by subtracting two points on the plane. Say $(1,1,\frac{-51}{5})$, $(1,2,\frac{-44}{5})$ and get the vector$(0,1,\frac{7}{5})$ on the plan. Then equation of a plan normal to the desired plan is $(y-1)+\frac{7}{5}(z-3)=0$. Is this correct?