Vectors: Can someone check my answer?

Question

1. Find an equation of the plane that contains the point $(2, 0, 3)$ and the line $x = −1 + t, y = t, z = −4 + 2t$. I plugged in $t=0$ to get $(-1,0,-4)$ and $t=1$ to get $(0,1,-2)$ $n1$ (vector from $(0,1,-2)$ to $(-1,0,-4)): (-1,-1,-2) n2$ (vector from $(0,1,-2)$ to $(2,0,3)): (2,-1,5) n1*n2=(-7,1,3)$ General equation of plane: $(-7,1,3)*(x-2,y-0,z-3) =-7x+y+3z=-5$

2. Find a plane through the points $P1(−2, 1, 4)$ and $P2(1, 0, 3)$, and perpendicular to the plane $4x−y+3z =2$ I found $v$ to equal $(3,-1,-1)$ and $n$ to equal $v*n2=(-4,-13,1)$ Then, $-4(x+2)-13(y-1)+1(z-4)=0 -4x-13y+z+1=0$

3. Find the distance between the given parallel planes: $−2x + y + z = 0 and 6x − 3y − 3z − 5 = 0$ I set $-2x+y+z=0, 6x-3y-3z-5=0 y=z=0$, so $-2x=0, x=0$ (passes through origin) Distance between $6x-3y-3z-5=0$ and origin $(0,0,0)$: abs$(6(0)-3(0)-3(0)-5)/\sqrt{y^2+(-3)^2+(-3)^2} =5/sqrt(54) =5/3sqrt(6) $

Answer 1

Let $A=(2,0,3)$ and the point on the line is $B(-1,0,-4)$ so $\vec {AB}=3 \vec i+7 \vec k$. the direction vector of the line is $\vec L=\vec i+ \vec j+ 2\vec k$. So the equation of the plane is $$u(x-2)+v(y-0)+w(z-3)=0~~~~(1)$$ where $\vec n=u \vec i+v \vec j+ w \vec k= \vec{AB} \times \vec L=-7 \vec i+ \vec j+ 3\vec k$. From (1), the equation of plane is $$-7(x-2)+1(y-0)+3(z-3)=0 \implies -7x+14+y+3z-9=0 \implies 7x -y-3z-5=0$$