Find the line through (7, 1, −6) that intersects and is perpendicular to the line x = −1 + t, y = −2 + t, z = −1 + t. (hint: if (x0, y0, z0) is the point of intersection, find its coordinates. enter your answers as a comma-separated list of equations.)

Step-by-step explanation:

We have been given the intersection coordinates:

(7,1,-6) and perpendicular to the line x=-1+t ,y=-2+t and z=-1+t

From the condition of perpendicularity we know:  

Now, we have given x=-1+t , y=-2+t and  z= -1+t

Now, substituting t in the given coordinates x,y and z we get:

And

 The answer is  

Step-by-step explanation: The vector equation of the line can be written as

The general equation of the plane that is perpendicular to this line will be

Since it contains the point (7,1,-6), so

Therefore, the plane x+y+z=2 contains the point (7,1,-6) and is perpendicular to the line.

Now, we will substitute the parametric equations of the line into the equation of the plane as follows

So, the parametric equation of the new line is

c-4 is the final answer! : )