Find the points of intersection of the line x = 3+2t, y =5+7t, z = -3+t, that is, I(t) = 3+2t, 5+7t, -3+t, with the coordinate planes.

The equations of the coordinate planes is given below:
$yz:x=0,zx:y=0,xy:z=0$
Consider in xy plane z=0.
Substitute 0 for z in the equation of line as given below,
${\displaystyle z=0\Rightarrow -3+t=0}$
t=3, now $x=3+2t$
At $t=3$,
$x=3+2\ast 3=3+6=9$
At t=3, $y=5+7\ast 3=5+21=26$
Point is (9,26,0)
Consider in yz plane x=0.
Substitute 0 for x in the equation of line as given below,
${\displaystyle x=0\Rightarrow 3+2t=0}$
2t=-3. now
At ${\displaystyle t=-\frac{3}{2}}$
${\displaystyle y=5+7\cdot (-\frac{3}{2})=\frac{10-21}{2}=-\frac{11}{2}}$
At ${\displaystyle t=-\frac{3}{2}}$
${\displaystyle z=-3-\frac{3}{2}=\frac{-6-3}{2}=-\frac{9}{2}}$
Point is ${\displaystyle (0,-\frac{11}{2},-\frac{9}{2})}$
Consider in zx plane y=0.
Substitute 0 for y in the equation of line as given below,
${\displaystyle y=0\Rightarrow 5+7t=0}$
7t=-5, now
At ${\displaystyle t=-\frac{5}{7}}$
${\displaystyle x=3+2\cdot (-\frac{5}{7})=\frac{21-10}{7}=\frac{11}{7}}$
At ${\displaystyle t=-\frac{5}{7}}$
${\displaystyle z=-3-\frac{5}{7}=-21-\frac{5}{7}=-\frac{26}{7}}$
Point is ${\displaystyle (\frac{11}{7},0,-\frac{26}{7})}$
Thus, the point of intersections on the coordinate planes are as given below:
xy:(9,26,0)
yz:${\displaystyle (0,\frac{11}{2},\frac{9}{2})}$
zx:${\displaystyle (\frac{11}{7},0,-\frac{26}{7})}$