Suppose we have plane such that $(0,-3,1)$ lies on the plane and is perpendicular to the planes $x+2y+3z=100$ and $3x-5y+4z=200$. What is the equation of this plane?

### attempt:

We know $\vec{n\_1}=(1,2,3)$ and $\vec{n\_2}= (3,-5,4)$ are the normals of the given planes. Since the two planes are perpendicular to the plane we want to find, then we can find some $\vec{v} =(a,b,c)$ on the plane such that $ \vec{v} \cdot \vec{n\_1} = \vec{v} \cdot \vec{n\_2} = 0$. These equations gives $a+2b +3c = 0$ and $3a-5b+4c=0$. Let $c = t$, and solving the system, we find that $\vec{v} = ( - \frac{23}{11} t, - \frac{5}{11} t, t )$ is the vector that lies on the plane. With $t = 11$, we have $\vec{v} = (-23,-5,11)$ lies on the plane. and this is the normal of the plane we want. Since point $(0,-3,1)$ is in the plane, then we have $\vec{r} = (x,y+3,z-1) \cdot \vec{v} = 0$ and so

$$ -23x - 5y - 15 + 11z - 11 = 0 \implies -23x-5y+11z-26 = 0$$

is the equation of the plane we want.

IS this correct?