Q: Parametric equation for the tangent curve 
Find parametric equations for the tangent line to the curve  with the
  parametric equations $x=t, y=t^2, z=t^3$ at the point $(1, 1, 1)$.

For the Solution I know the method of solving it. But have a small problem in the procedure:  
The parameter corresponding to the point $(1, 1, 1)$ is $t = 1$, so 
And we have that $r(t)=(t,t^2,t^3)$
Thus $r'(1)=(1,2,3)$ 
And my doubt is after this step why do we write the parametric equation as  
$x=1+t, y=1+2t, z=1+3t$? 

A: The (direction) vector $\mathbf{v}=(1,2,3)$ is the direction of your curve (and therefore the tangent line) at this (position) vector $\mathbf{p}=(1,1,1)$. 
Then you can use that the (vector-valued) equation of the line through $p$ with direction $\mathbf{v}$ is given by 
$$(x(t),y(t),z(t))=L(t)=\mathbf{p}+\mathbf{v}t=(1,1,1)+(1,2,3)t=(1+t,1+2t,1+3t)$$
from which you can read off the conclusion you desire.
