Finding a perpendicular plane Question
Plane $B$ contains the points $(4, 2, 1)$ and $(4, 1, -6)$, and it is perpendicular to the plane $7x+9y+4z=18$ . What is the equation of the plane $B$?
What I tried:
I know that to find the equation of plane $B$ I need a point on the plane and a perpendicular vector. The point part is already covered $(4, 1, -6)$, and the cross product of $\langle4-4, 2-1, 1-(-6)\rangle$ and $\langle7, 9, 4\rangle$ should give me the perpendicular vector; then it's just a matter of plugging the numbers in the formula.  However, this approach did not work and I don't know what I am doing wrong. 

Your approach is both correct and simple.
Your plane has to be parallel to both the vectors
$$\begin{pmatrix}4\\2\\1\end{pmatrix}-\begin{pmatrix}4\\1\\-6\end{pmatrix}=\begin{pmatrix}0\\1\\7\end{pmatrix}
\qquad\text{and}\qquad
\begin{pmatrix}7\\9\\4\end{pmatrix}$$
so its normal vector can be computed as
$$\begin{pmatrix}0\\1\\7\end{pmatrix}\times
\begin{pmatrix}7\\9\\4\end{pmatrix}=
\begin{pmatrix}-59\\49\\-7\end{pmatrix}$$
Plug in one of your two points
$$-59\cdot4+49\cdot2-7\cdot1=-145$$
so if you also get rid of some minus signs, the equation of the plane is
$$59\,x-49\,y+7\,z=145$$
To verify the required properties, you can plug in both your points and see that they satisfy the equation, and you can also compute the dot product
$$\left<\begin{pmatrix}59\\-49\\7\end{pmatrix},
\begin{pmatrix}7\\9\\4\end{pmatrix}\right>=
59\cdot7-49\cdot9+7\cdot4=0$$
which proves that the normals are perpendicular, therefore the planes are perpendicular.
