**find vector and parametric equations for the line passing through point (6,-5,2) and perpendicular to the plane x+3y+z=5**

**Answer:**

To find a direction vector, notice that since the line is to be perpendicular to the given plane, the normal vector [1,3,1] of the plane can serve as the direction vector for the line. So we select the fixed point as (6,-5,2) and the direction vector for the line is v = [ 1,3,1]. Then the vector equation of the line is r = r[0] + t v . The parametric equations are

x=6+t,y=−5+3t,z=2+t