Find the equation of the plane through the pt (6, 3, 2) and perpendicular to the vector (-2, 1, 5)

Also, is the vector the same as the unit vector since it's perpendicular to the plane? Thanks

Note that since the plane is perpendicular to <-2, 1, 5>, the plane has a normal vector of <-2, 1, 5>.

Since the components of the normal vector are the coefficients of x, y, and z in the equation of the plane, we see that the plane has equation:
-2x + y + 5z = c, for some c.

Since (6, 3, 2) lies on the plane:
-2(6) + 3 + 5(2) = c ==> c = 1.

Therefore, an equation of the plane is -2x + y + 5z = 1.

I hope this helps!

P(x, y, z) is any point on the plane. By dot product, you have the answer.
-2(x-6) + (y-3) + 5(z-2) = 0