You were asked to find the vector and Cartesian forms for a line.
You gave, as an answer, 14x +2y+ 7z= 79. In the first place you should recognize that this is NOT the equation of ANY line in three dimensions! A single equation is not the equation of a line- a single equation reduces the dimension by one. In the plane that is 2- 1= 1 so an equation can describe a line but in 3 dimensions, a singe equation reduces to 3- 1= 2, a plane. The problem you were given told you that x- 7y+ 2z= 4 is the equation of a plane. I am puzzled why you would think 14x+ 2y+ 7z= 79 would be the equation of a line!
If you were expected to be able to do a problem like this, I would think you were expected to know that any line perpendicular to the plane x- 7y+ 2z= 4, would have to have parametric for x= t+ a, y= -7t+ b, z= 2t+ c. Since with t= 0 that gives x= a, y= b, z= c. If we want the line to pass through the point (4, 4, 2) we clearly have to take a= 4, b= 4, c= 2 so x= t+ 4, y= -7t+ 4, z= 2t+ 2.
Of course, that can be written in vector form (t+ 4)i+ (-7t+ 4)j+ (2t+ 2)k (or as (ti- 7tj+ 2tk)+ (4i+ 4j+ 2k). Since x= t+ 4, t= x- 4, since y= -7+ 4, t= (y-4)/(-7), and since z= 2t+ 2. t= (z- 2)/2. Setting those forms of t equal we get one "Cartesian" form, the "symmetric" form, x- 4= (y- 4)/(-7)= (z- 2)/2.
