Find the equation of a plane that goes through the intersection of the planes $x_1+5x_2+x_3-5=0$ and $x_1-3x_3-2x_3=0$ and contains the point $(1,1,0)$ .
I calculated the cross product of the given plane to obtain the intersection and it gave me $-7x_1+3x_2-8x_3$, but I'm stuck there.
On
You need two vectors for the plane you want, the one you got is$(-7,3,-8)$, so you need to find another one, choose a point P which lies both on $x_1+5x_2+x_3-5=0$ and $x_1-3x_2-2x_3=0$ (I assume you had a type mistake), and the vector from P to $(1,1,0)$ will be the second vector, do cross product then you will have the normal vector of the plane you want, $(1,1,0)$ must be contained so you can get the constant from $(1,1,0)$.
Given $$\alpha:x_1+5x_2+x_3-5=0,\;\beta:x_1-3x_2-2x_3=0$$ Any linear combination of the two planes passes through the intersection line $$\lambda(x_1+5x_2+x_3-5)+\mu(x_1-3x_2-2x_3)=0\tag{1}$$ plug the coordinates of the point $(1,1,0)$ $$\lambda(1+5+0-5)+\mu(1-3-0)=0\to \lambda=2\mu$$ substitute in $(1)$ $$2\mu(x_1+5x_2+x_3-5)+\mu(x_1-3x_2-2x_3)=0$$ $$2(x_1+5x_2+x_3-5)+(x_1-3x_2-2x_3)=0\to 3x_1+7x_2-10=0$$