Linear Algebra - point not on plane I have this as a practice problem for my linear algebra course. How should I approach this problem? Hints would be useful.

 A: $\textbf{Tricky solution}$ Let's try solve this question quickly: guess that $(1,1,-4)$ and $(6,2,4)$ lay on the plane which we are looking for, then $(1,1,-4)+(6,2,4)=(7,3,0)$ also lay on the plane and note that this point lay on plane:
$$3x-7y+z=0$$
Now note that all point except $(2,1,1)$ lay on this plane.
$\textbf{Formal solution}$ Formally you should get every three point, calculate the equation of plane $ax+by+cz=0$ on which they lay by solving system of equations:
$$x_1a+y_1b+z_1c=0$$
$$x_2a+y_2b+z_2c=0$$
$$x_3a+y_3b+z_3c=0$$
Then check if three other points lay on this plane.
A: make a $3 \times 7$ matrix of you seven points making a column for each point. row reduce the matrix. you will see that the vectors $3, 4, 5, 6$ and $7$ are linear combinations of the first two. so the $(2,1,1)$ is the odd one out.
A: Try crossing two of them and see if the result is perpendicular to all but one others. Start with the first 2 vectors:
$$
   (1,1,-4) \times (3,5,-26) = (-6,14,2)
$$
Automatically $(-6,14,2)$ is orthogonal to the first two vectors.
Now check the other 
dot products:
$$
\begin{align}
   (2,1,1)\cdot(-6,14,2) & = -12+14+2 & \ne 0,\\
   (5,6,-27)\cdot(-6,14,2) & = -30+84-54 & = 0,\\
   (4,1,5)\cdot(-6,14,2) & = -24+14+10 & = 0,\\
   (5,-2,29)\cdot(-6,14,2) & = -30-28+58 & =0, \\
   (6,2,4)\cdot(-6,14,2) & = -36+28+8 & = 0.
\end{align}
$$
So the only point not on the plane $-6x+14y+2z=0$ is the third point $(2,1,1)$.
