use the result of Exercise 35 to find parametric equations for the line segment connecting point P to point Q.

$P(1,7,3),Q(-1,-2,5)$

$\text{The points}P(1,7,3)\text{and}Q(-1,-2,5)\text{.}$

$\text{The point are}P(1,7,3)\text{and}Q(-1,-2,5)\text{.}$

$\overrightarrow{PQ}=(-1-1,-2-7,5-3)$

$\overrightarrow{PQ}=(-2,-9,2)$

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.

$\text{Here}P(1,7,3)\text{and}\overrightarrow{PQ}=d=(-2,-9,2)\text{then the equation is,}$

$r\left(t\right)=(1,7,3)+t(-2,-9,2)$

$r\left(t\right)=(1-2t,7-9t,3+2t)$

$\text{The equation is}r\left(t\right)=(1-2t,7-9t,3+2t)\text{.}$

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

$\text{Then,}r\left(t\right)=\left(x\right(t),y(t),z(t\left)\right)=(1-2t,7-9t,3+2t)$

$\text{The parametric equations are}x\left(t\right)=1-2t,y\left(t\right)=7-9t,z\left(t\right)=3+2t$

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 $0\le t\le 1$

Thus, the parametric equations are $x\left(t\right)=1-2t,y\left(t\right)=7-9t,z\left(t\right)=3+2t$where $0\le t\le 1$

$\text{Therefore, the answer is}x\left(t\right)=1-2t,y\left(t\right)=7-9t,z\left(t\right)=3+2t\text{.}$