91Ó°ÊÓ

$\text{The points}P(0,0,0)\text{and}Q(1,2,3)\text{.}$

$\text{The point are}P(0,0,0)\text{and}Q(1,2,3)\text{.}$

$\stackrel{→}{PQ}=(1-0,2-0,3-0)$

$\stackrel{→}{PQ}=(1,2,3)$

The formula to find the line $\mathcal{L}$equation is as follows, $r\left(t\right)=P_0+td$Where, $P_0$is the point and $d$is the direction vector.

$\text{Here}P(0,0,0)\text{and}\stackrel{→}{PQ}=d=(1,2,3)\text{then the equation is,}$

$r\left(t\right)=(0,0,0)+t(1,2,3)$

$r\left(t\right)=(0+t,0+2t,0+3t)$

$\text{The equation is}r\left(t\right)=(t,2t,3t)\text{.}$

$\text{The vector function}r\left(t\right)\text{in three -dimensional plane represents}r\left(t\right)=\left(x\right(t),y(t),z(t\left)\right)\text{.}$

$\text{Then,}r\left(t\right)=\left(x\right(t),y(t),z(t\left)\right)=(t,2t,3t)$

$\text{Thus the parametric equations are}x\left(t\right)=t,y\left(t\right)=2t,z\left(t\right)=3t\text{.}$

The restriction for the parameter $t$so that the result parametrizes the segment P to point Q is given from 0 to 1 that is from $0≤t≤1$.

Thus, the parametric equations are $x\left(t\right)=t,y\left(t\right)=2t,z\left(t\right)=3t$where $0≤t≤1$

$\text{Therefore, the answer is}x\left(t\right)=t,y\left(t\right)=2t,z\left(t\right)=3t,0≤t≤1\text{.}$