Answer
(a) $x=2+t,y=4-t,z=6+3t$
(b) $(0,6,0)$, $(6,0,18)$, $(0,6,0)$.
Work Step by Step
(a) Equation of the line is given by $r=r_0+tv$
$r_0=(2,4,6)$ and $v= \lt 1,-1,3 \gt$
Parametric equations are defined by:
$x=x_0+at$, $y=y_0=bt$ and $z=z_0+ct$
Thus, the parametric equations are:
$x=2+t,y=4-t,z=6+3t$
(b) From part (a), we have
$x=2+t,y=4-t,z=6+3t$
For the intersection with the $xy-plane$, set $z=0$ and solve for $x$ and $y$.
Therefore, the point of intersection with the $xy-plane$ is $(0,6,0)$.
For the intersection with the $xz-plane$, set $y=0$ and solve for $x$ and $z$.
Therefore, the point of intersection with the $xz-plane$ is $(6,0,18)$
For the intersection with the $yz-plane$, set $x=0$ and solve for $y$ and $z$.
Therefore, the point of intersection with the $yz-plane$ is $(0,6,0)$.
Hence, $(0,6,0)$, $(6,0,18)$, $(0,6,0)$.
