Answer:
The parametric equations of line are x=-8-t, y=5+8t and z=2+t.
Explanation:
It is given that the line passes through the point (−8, 5, 2) and is perpendicular to the plane given by −x + 8y + z = 5.
The coordinate of point are (−8, 5, 2).
Normal vector = <-1,8,1>
The position vector of a line is
![\overrightarrow{r}=(x_0,y_0,z_0)+t<a,b,c>](https://img.qammunity.org/2020/formulas/mathematics/college/e8pnbrf4h0hvfe87obhob3nh7ryt8sgzar.png)
then the parametric equations of line are
.
Where, (x_0,y_0,z_0) are the coordinate of point and <a,b,c> is normal vector.
The position vector of given line is
![\overrightarrow{r}=(-8,5,2)+t<-1,8,1>](https://img.qammunity.org/2020/formulas/mathematics/college/9qola0qie8mjatfuwgrici5we2vty1j8yb.png)
The parametric equations of line are
![x=-8-t](https://img.qammunity.org/2020/formulas/mathematics/college/bpjc9kcxctl9h9xo525qtfynehxd8yhjk7.png)
![y=5+8t](https://img.qammunity.org/2020/formulas/mathematics/college/7fpm1hfj2kn9b4drwsiqbr9vgs7v4fsn5d.png)
![z=2+t](https://img.qammunity.org/2020/formulas/mathematics/college/xvpkhjzkyvec71jn01rf44oqglrtgjloqz.png)
Therefore the parametric equations of line are x=-8-t, y=5+8t and z=2+t.