Find the value of $c$ if the vectors lie in the same plane So I have three vectors $\vec a =(2, -2, c), \vec b=(1, 2, 3) $ and $\vec c= (4, 2, 2)$ which are all on the same plane that passes through the origin and I want to find the value of $c$.
So I'm just confused how to find the value of $c$.
I've tried using two methods but none of them work out to the final answer which seems to be $c = -4$.
First method: 
I first tried using the fact that three vectors lie on the same plane iff $\vec a.(\vec b×\vec c) =0$ as seen in this answer here. So in this case my working is as follows. 
$$ \overrightarrow b × \overrightarrow c = (4-6)\overrightarrow i +(2-12) \overrightarrow j + (2-8) \overrightarrow k  $$
$$ \overrightarrow b × \overrightarrow c = -2\overrightarrow i -10 \overrightarrow j -6 \overrightarrow k  $$
$$ \overrightarrow a⋅(\overrightarrow b × \overrightarrow c) = (2*-2)  + (-2 *-10) + (c* -6) = 0  $$
$$ \overrightarrow a⋅(\overrightarrow b × \overrightarrow c) = -4  + 20 -6c = 0  $$
$$ -6c = -16  $$
$$ c = 8/3  $$
Second method: 
I tried finding the equation of the plane in 3d space using the three vectors $\overrightarrow P$ =(0, 0, 0) since the plane passes through the origin, $\overrightarrow Q$=(1, 2, 3) and $\overrightarrow R$= (4, 2, 2) and then substituting the point (2, -2, c) into the equation of the plane.
$$ \overrightarrow {PQ} = \overrightarrow Q - \overrightarrow P = (1,2,3) $$
$$ \overrightarrow {PR} = \overrightarrow R - \overrightarrow P = (4,2,2) $$
Next I crossed $\overrightarrow {PQ} × \overrightarrow {PR}$ to find the tangent vector which is the coefficients for the equation of the plane where the tangent vector = (a,b,c) and the equation of the plane is $a(x-x0) + b(y-y0) + c(z-z0) = 0$.
$$\overrightarrow {PQ} × \overrightarrow {PR}= -2\overrightarrow i -10 \overrightarrow j -6 \overrightarrow k $$
Thus, the equation for the tangent plane is:
$$-2(x-x0) -10(y-y0) + -6(z-z0) = 0$$
Substituting any random easy point on the plane (0,0,0).
$$-2x -10y -6z = 0$$
Substituting the point in question that must be on the plane(2, -2, c).
$$-2(2) -10(-2) -6c = 0$$
$$-2(2) -10(-2) -6c = 0$$
$$-6c = -16$$
$$c = 8/3$$
So I was just wondering if I had some kind of error in both my methods because the answer is c=-4 or if their answer was wrong and it seems like the first method seems to be a lot more efficient.
 A: Okay @AlexeyBurdin has pointed out my mistake essentially I got the cross product wrong in both methods, but they should both work, I'm just going to put this correction up and what the subsequent answer should be for anyone else who runs into this problem again.
The reason is the for the matrix $$
  A=\begin{pmatrix}
    a & b & c\\
    d & e & f\\
    g & h & i\\
  \end{pmatrix}
$$ |A| = a(ei − fh) − b(di − fg) + c(dh − eg)
And the minus in the middle is what I forgot about which lead to the rest of my workings being wrong and finding the determinant of a 3 x 3 matrix is part of what the cross product is.
First method: 
$$ \overrightarrow b × \overrightarrow c = (4-6)\overrightarrow i - (2-12) \overrightarrow j + (2-8) \overrightarrow k  $$
$$ \overrightarrow b × \overrightarrow c = -2\overrightarrow i +10 \overrightarrow j -6 \overrightarrow k  $$
$$ \overrightarrow a⋅(\overrightarrow b × \overrightarrow c) = (2*-2)  + (-2 *10) + (c* -6) = 0  $$
$$ \overrightarrow a⋅(\overrightarrow b × \overrightarrow c) = -4  - 20 -6c = 0  $$
$$ -6c = 24  $$
$$ c = -4  $$
Second method: 
$$ \overrightarrow {PQ} = \overrightarrow Q - \overrightarrow P = (1,2,3) $$
$$ \overrightarrow {PR} = \overrightarrow R - \overrightarrow P = (4,2,2) $$
Next I crossed $\overrightarrow {PQ} × \overrightarrow {PR}$ to find the tangent vector which is the coefficients for the equation of the plane where the tangent vector = (a,b,c) and the equation of the plane is $a(x-x0) + b(y-y0) + c(z-z0) = 0$.
$$\overrightarrow {PQ} × \overrightarrow {PR}= -2\overrightarrow i +10 \overrightarrow j -6 \overrightarrow k $$
Thus, the equation for the tangent plane is:
$$-2(x-x0) +10(y-y0) + -6(z-z0) = 0$$
Substituting any random easy point on the plane (0,0,0).
$$-2x +10y -6z = 0$$
Substituting the point in question that must be on the plane(2, -2, c).
$$-2(2) +10(-2) -6c = 0$$
$$-2(2) +10(-2) -6c = 0$$
$$-2(2) +10(-2) -6c = 0$$
$$-24-6c = 0$$
$$-6c = 24$$
$$c = -4$$
