91Ó°ÊÓ

$\text{The points}P(0,2,3)\text{and}Q(4,5,-1)\text{.}$

$\text{The point are}P(0,2,3)\text{and}Q(4,5,-1)$

$\stackrel{→}{PQ}=(4-0,5-2,-1-3)$

$\stackrel{→}{PQ}=(4,3,-4)$

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 dis the direction vector.

$\text{Here}P(0,2,3)\text{and}\stackrel{→}{PQ}=d=(4,3,-4)\text{then the equation is,}$

$r\left(t\right)=(0+4t,2+3t,3-4t)$

The equation is written as follows,

$r\left(t\right)=(4t,2+3t,3-4t)$

The vector function $r\left(t\right)$in three -dimensional plane represents $r\left(t\right)=\left(x\right(t),y(t),z(t\left)\right)$. Then, $r\left(t\right)=\left(x\right(t),y(t),z(t\left)\right)=(4t,2+3t,3-4t$)

Thus the parametric equations are $x\left(t\right)=4t,y\left(t\right)=2+3t,z\left(t\right)=3-4t$.

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)=4t,y\left(t\right)=2+3t,z\left(t\right)=3-4t$$ where $0≤t≤1$.

$\text{Therefore, the answer is}x\left(t\right)=4t,y\left(t\right)=2+3t,z\left(t\right)=3-4t,0≤t≤1\text{.}$