Answer:
They are not.
Step-by-step explanation:
Let's first calculate the vectors joining the first point to the second and the first to the third, they will be useful later.
[tex]\vec v_1_2 = \vec P_2 -\vec P_1 = <3; -7; -7>\\\vec v_1_3= \vec P_2 -\vec P_1 = <6; -14; -13>[/tex]
At this point we have at least three options:
Write down the line between P1 and P2 in vector form, and then see if we can fit the third point in there.
Calculate the cross product between the two, if it's non-zero the three point are not in a line.
Try to determine the plane passing for the 3 points: if we find one they are not in a line.
Option 1:
The vector form of the lne between the firs two points is [tex]<6+3t;9-7t;7-7t>[/tex]. Let's check the coordinates of the third point:
[tex]x_3:\ 6+3t= 12 \rightarrow t=2\\y_3:\ 9-7t = -5 \rightarrow t =2\\7z_3:\ 7-7t=-6 \rightarrow t = 13/7\ne 2[/tex]
So the three points are not in a line.
Option 2:
Let's calculate the cross product of the two vectors we found at the beginning.
[tex]\vec v_1_2 \times \vec v_1_3 = det \left[\begin{array}{ccc}\hat i&\hat j&\hat k\\3&-7&-7\\6&-14&-13\end{array}\right] = \hat i [-7(-13)-(-14)(-7)] - \hat j[3(-13)-6(-7)]+\hat k [3(-14)-6(-7)] = 7\hat i -3 \hat j + 0 \hat k \ne 0[/tex]
Since the result is not the null vector, the three points are not in a line.
Option 3:
Let's write the plane (in the [tex]z= ax+by+c[/tex] form that contain the three points. If the thre points are in a line, means that we can find only two of these parameters.
[tex]P1: 7= 6a+9b+c\\P2: 0= 9a+6b+c\\P3: -6= 12a-5b +c[/tex]
Since you can determine a, b and c ( I ended up with [tex]a=-\frac {10}3; b=-\frac18; c= \frac{123}8[/tex] but I probably messed something in there) the thre points are not in a line
