Answer
$4x-2y-z=14$
Work Step by Step
Step 1. Given the line equations $x=2+2t, y=4t, z=-6$, we can find two points on the line by letting $t=0, 1$ which gives: for $t=0$, $Q(2,0,-6)$ and for $t=1$, $R(4,4,-6)$
Step 2. Establish two vectors on the plane: $\vec u=\vec {PQ}=\langle 2-5,0-3,-6-0 \rangle=\langle -3,-3,-6 \rangle$ and $\vec v=\vec {PR}=\langle 4-5,4-3,-6-0 \rangle=\langle -1,1,-6 \rangle$
Step 3. Find a normal to the plane with the cross product of the above two vectors: $\vec u\times\vec v=\langle (18+6),(6-18),(-3-3) \rangle=\langle 24,-12,-6 \rangle$. Simplify this vector as $\vec n=\langle 4,-2,-1 \rangle$
Step 4. The general equation for a plane containing point $P(x_0,y_0,z_0)$ and having a normal $\vec n=\langle a,b,c \rangle$ is $a(x-x_0)+b(y-y_0)+c(z-z_0)=0$
Step 5. The equation of the plane in the problem can be written as: $4(x-5)-2(y-3)-(z-0)=0$ or $4x-2y-z=14$
