Equation of a intersected plane in space

Question

Find the equation of the plane that passes through the points $(5,4,1)$ and $(5,1,-6)$ and is perpendicular to the plane $7x+8y+2z=20$.

First I found the vector of the two points
$$(5-5,1-4,-6-1)= <0,-3,-7>=\vec v$$ Knowing that the the plane equ. gives us $<7,8,2>$ we can then cross the two vectors and get and new vector, orthogonal to our original plane $$\vec u=<50,49,21>$$ Now we plug in $\vec v$ and $\vec u$ into the general equation for a plane $$50(x-0)+49(y+3)+21(z-7)=0$$ $$50x+49y+21z+294=0 $$ Now somewhere I went wrong, I went back through my cross product and almost convinced myself that I should reverse the matrix and try again but that would just result in the negative of what I already have.

Answer 1

First of all you have a sign issue, as mentioned in a different answer$$\vec u=<50,-49,21>$$ Then your mistake is that you inserted $\vec v$ in the equation of the plane. You should instead insert either of the two points in that equation. There is no guarantee that the vector you get from subtracting coordinates of two points is in the plane. In this case it is not.

So the equation of the plane is $$50x-49y+21z=d$$ Plug in any of the initial points, and you get $d$. Jut to check, plug in the other point, and you should obtain the same $d$.

Answer 2

I think, it should be $$\vec{u}(50,-49,21).$$

Let $\vec{u}(a,b,c).$

Thus, $$3b+7c=0$$ and $$7a+8b+2c=0.$$

I hope, now it's clear.