Find a plane which is orthogonal to P1 and P2 and state its general equation I'm having a bit of trouble approaching this problem
Let P1 and P2 be planes with the general equations
P1 = -2x + y - 4z =2
P2 = x + 2y = 7
I understand that the unit vectors are (-2, 1, -4) & (1, 2, 0)
And I know that the unit vector of the general equation of P3 must dot product with the 2 above unit vectors to give 0.
Where do I go from the information I have?
 A: You don't have to use the cross product if you're not comfortable with it. You're looking for a normal vector, and as you say, it must dot product with the two given vectors to produce $0$. That is, if $(a, b, c)$ is our normal vector, then
\begin{align*}
0 &= (-2, 1, 4) \cdot (a, b, c) = -2a + b + 4c \\
0 &= (1, 2, 0) \cdot (a, b, c) = a + 2b.
\end{align*}
We can solve this system of linear equations. Note that these equations are homogeneous, meaning that $(a, b, c) = (0, 0, 0)$ is definitely a solution (though useless for our purposes). Also, there are only two equations and three variables, so we can expect infinitely many solutions to the system. This should make sense, as we can scale any non-zero normal vector as we like, and still get a normal vector.
As an augmented matrix, we get
\begin{align*}\left(\begin{array}{ccc|c}-2 & 1 & 4 & 0 \\ 1 & 2 & 0 & 0\end{array}\right) &\sim \left(\begin{array}{ccc|c}1 & 2 & 0 & 0 \\ -2 & 1 & 4 & 0\end{array}\right) \\
&\sim \left(\begin{array}{ccc|c}1 & 2 & 0 & 0 \\ 0 & 5 & 4 & 0\end{array}\right) \\
&\sim \left(\begin{array}{ccc|c}1 & 2 & 0 & 0 \\ 0 & 1 & 4/5 & 0\end{array}\right).
\end{align*}
Note that the third column has no pivot, so $c$ shall be our free variable. The second row implies
$$b + \frac{4}{5}c = 0 \implies b = -\frac{4}{5}c.$$
The first row implies
$$a + 2b = 0 \implies a = -2b = \frac{8}{5}c.$$
Thus, the general solution is
$$(a, b, c) = \left(\frac{8}{5}c, -\frac{4}{5}c, c\right) = \frac{c}{5}(8, -4, 1).$$
We can choose any $c \neq 0$ to obtain a valid normal direction. I suggest choosing $c = 5$, so that you get all integers (for the sake of tidiness). This gives us the normal direction $(8, -4, 5)$. So, any plane of the form
$$(8, -4, 5) \cdot (x, y, z) = ?$$
will do the trick, e.g.
$$8x - 4y + 5z = 0.$$
A: Hint
If $u_1= ^t(-2, 1, -4)$, $u_2= ^t(1, 2, 0)$ and $u=u_ 1 \times u_2 = ^t(u_x,u_y,u_z)$ is the cross product of $u_1,u_2$, then the equation of the plane $P$ passing through the origin and orthogonal to the planes $P_1, P_2$ is 
$$u_x x+u_y y+u_z z =0$$
And the equation of a plane parallel to $P$ is
$$u_x x+u_y y+u_z z =\lambda$$
With that, you have all the planes you're looking for!
A: Let $u_1=\langle -2,1,-4\rangle$ and $u_2=\langle 1,2,0\rangle$.
The cross product $u_1\times u_2$ is defined as follows:
$$u_3=u_1\times u_2=
\begin{vmatrix} 
\hat x & \hat y & \hat z \\
 -2 & 1 & -4 \\ 
1 & 2 & 0 
\end{vmatrix}$$
The resulting vector $u_3$ is orthogonal to the two normal vectors you listed. Can you find the determinant to get $u_3$?
