Answer:
![r=(0,11,-8)+(4,-3,7)t](https://tex.z-dn.net/?f=r%3D%280%2C11%2C-8%29%2B%284%2C-3%2C7%29t)
x=4t
y=11-3t
z=-8+7t
Step-by-step explanation:
The line is parallel to x=-1+4t, y=6-3t, z=3+7t.
Two lines are parallel if they have the same direction, and in the parametric form, the direction of a line is always the vector of constants that multiply t (or the parameter). So in this case, the direction is (4,-3,7); this is also the direction of the missing line because they are parallel.
The vector equation of a line is given by:
![r=r_{0} +tv](https://tex.z-dn.net/?f=r%3Dr_%7B0%7D%20%2Btv)
where v is the direction vector, and
is a point of the line.
So, for this case, the line pass for the point (0, 11, −8):
![r_{0}=(0,11,-8)](https://tex.z-dn.net/?f=r_%7B0%7D%3D%280%2C11%2C-8%29)
With the direction, the vector equation is:
![r=(0,11,-8)+(4,-3,7)t](https://tex.z-dn.net/?f=r%3D%280%2C11%2C-8%29%2B%284%2C-3%2C7%29t)
The parametric equations are just the simplification of the vector equation:
x=4t
y=11-3t
z=-8+7t