Answer:
(1+t, 2+2t, -1+6t); C is on this line.
Step-by-step explanation:
Let's find the vector equation of the line.
Given information:
Point A is (1, 2, -1)
Point B is (2, 4, 5)
Point C is (-5, -10, -37)
The difference between the two first given points (A, B) is:
B-A=(2, 4, 5)-(1, 2, -1)=(2-1, 4-2, 5-(-1))=(1, 2, 6) which is called the direction vector, so the equation is:
(x, y, z)=(1, 2, -1) + t*(1, 2, 6)=(1+t, 2+2t, -1+6t)
Now let's find if point C is on the line:
(x, y, z)=(1, 2, -1) + t*(1, 2, 6)
(-5, -10, -37)=(1, 2, -1) + t*(1, 2, 6)
(-5, -10, -37)-(1, 2, -1)=t*(1, 2, 6)
(-5-1, -10-2, -37+1)=t*(1, 2, 6)
(-6, -12, -36)=t*(1, 2, 6) from where we can obtain:
1*t=-6 --> t=-6
2*t=-12 --> t=-6
6*t=-36 --> t=-6
In conclusion, because from all the equations we obtained t=-6, then point C is on this line.
