(a) i. See comment...
(a) ii. We have
[tex]\vec p = 6\,\vec\imath - 3\,\vec\jmath + 9\,\vec k[/tex]
and
[tex]\vec q = 3\,\vec\imath - 6\,\vec\jmath - 3\,\vec k[/tex]
(b) We can parameterize the line segment from P to Q by the vector function
[tex]\vec r(t) = (1-t) \vec p + t \vec q = (6-3t) \, \vec\imath + (-3-3t) \,\vec\jmath + (9-6t) \,\vec k[/tex]
with [tex]0\le t\le1[/tex]. The point R is located 1/3 of the way along this line segment, i.e. at [tex]t=\frac13[/tex], so its position vector is
[tex]\vec r\left(\dfrac13\right) = 5\,\vec\imath -4\,\vec\jmath + 7\,\vec k[/tex]
and hence its coordinates are [tex]\boxed{(5,-4,7)}[/tex].