Answer
$(t^2+t)i+(t^3+1)j+(t^4-1)k$
Work Step by Step
As we are given that $a(t)=2i+6tj+12t^2k$
Since, $v(t)=\int a(t)$ and $r(t)=\int v(t)$
Thus, $v(t)=\int (2i+6tj+12t^2k) dt$
$\implies v(t)=(2t+1)i+3t^2j+4t^3k$
Now, $r(t)=\int [(2t+1)i+3t^2j+4t^3k] dt=(t^2+t)i+t^3j+t^4k+c$
Here, $c$ represents a constant of integration.
That is, $r(0)=j-k \implies c=j-k$
or, $r(t)=(t^2+t)i+t^3j+t^4k+j-k$
Hence, $r(t)=(t^2+t)i+(t^3+1)j+(t^4-1)k$
