Answer
(a) The point $(2,1,-3)$ lies on the line.
(b) The point $(8,5,-11)$ lies on the line.
Work Step by Step
$(x,y,z) = (2,1,-3)+n(3,2,-4)$
(a) If $(2,1,-3)$ lies on the line, then there is a real number $n$ such that $(2,1,-3)+n(3,2,-4) = (2,1,-3)$
$x$: If $2+n(3) = 2$, then $n = 0$
$y$: If $1+n(2) = 1$, then $n = 0$
$z$: If $-3+n(-4) = -3$, then $n = 0$
Since the required value of $n$ is the same for $x,y,$ and $z$, the point $(2,1,-3)$ lies on the line.
(b) If $(8,5,-11)$ lies on the line, then there is a real number $n$ such that $(2,1,-3)+n(3,2,-4) = (8,5,-11)$
$x$: If $2+n(3) = 8$, then $n = 2$
$y$: If $1+n(2) = 5$, then $n = 2$
$z$: If $-3+n(-4) = -11$, then $n = 2$
Since the required value of $n$ is the same for $x,y,$ and $z$, the point $(8,5,-11)$ lies on the line.
