Answer
$r=(-\dfrac{t^{2}}{2}+1) {i}+(-\dfrac{t^{2}}{2}+2) {j}+(-\dfrac{t^{2}}{2}+3) {k}$
Work Step by Step
Given: $ \dfrac{d r}{d t} =-t{i}-t {j}-t {k}$ and $r(0) =i+2 j+3 {k} $
$r=\int(-t {i}-t {j}-t {k}) d t \\
=-\dfrac{t^{2}}{2} {i}-\dfrac{t^{2}}{2} {j}-\dfrac{t^{2}}{2} {k}+C$
Since $r(0) =i+2j+3k$, then
$r(0)=0 {i}-0 j-0k+C$
or, $={i}+2 {j}+3 {k} $
or, $C=i+2 j+3 k$
Hence
$r=(-\dfrac{t^{2}}{2}+1) {i}+(-\dfrac{t^{2}}{2}+2) {j}+(-\dfrac{t^{2}}{2}+3) {k}$
