Question

The plane 6x-2y+z=11 contains the line (x-1)=((y+1)/2)= ((z-3)/k). find k.

Answer 1

If the line lies in the given plane, its "direction vector" will be perpendicular to the plane's normal.

Do you know the plane's normal vector ?

Answer 2

normal vector would be <6,-2,1>

Answer 3

@phi

Answer 4

yes on the normal vector

you can write the equation of the line as
P + t V
where P is a point on the line (here (1,-1,3) )
and V is a direction vector = <1,2,k>

the dot product of V with the normal must give 0 (they are perpendicular vectors)

Answer 5

you want \[ <6,-2,1> \cdot <1,2,k>= 0 \\ 6-4+k=0 \\ k = -2 \]