Determine which points are on a line containing the point (-3,1,1) and parallel to the vector (1,-2,3)

Let $P = (-3,1,1)$ be a point in $\mathbb{R}^3$ and let $L$ be a line containing $P$ and parallel to the vector $(1,-2,3)$ . Determine which of the following points are also on $L$ .

$(0,0,0)$ ;
$(2,-1,4)$ ;
$(-2,-1,4)$ ;
$(-4,3,-2)$ ;
$(2,-9,16)$ .

Given a point $P= (-3,1,1)$ and a vector $A = (1,-2,3)$ the line containing $P$ in the direction of $A$ is given by

$L(P;A) = L((-3,1,1);(1,-2,3)) = \{ (-3,1,1) + t(1,-2,3) \mid t \in \mathbb{R} \} = \{(-3+t, 1-2t, 1+3t) \mid t \in \mathbb{R} \}.$

If $(0,0,0)$ were on $L$ then we must have some $t \in \mathbb{R}$ such that
$-3+t = 0, \qquad 1-2t = 0, \qquad 1+3t = 0.$

From the first equation, this requires $t = 3$ , but then neither of the other two equations can hold. Hence, there is no $t \in \mathbb{R}$ such that $(0,0,0) = (-3+t,1-2t,1+3t)$ so $(0,0,0)$ is not on the line.
If $(2,-1,4)$ were on $L$ then we must have some $t \in \mathbb{R}$ such that
$-3+t = 2, \qquad 1-2t = -1, \qquad 1+3t = 4.$

From the first equation, this requires $t = 5$ , but then neither of the other two equations can hold. Hence, there is no $t \in \mathbb{R}$ such that $(2,-1,4) = (-3+t,1-2t,1+3t)$ so $(2,-1,4)$ is not on the line.
If $(-2,-1,4)$ is on $L$ then we must have some $t \in \mathbb{R}$ such that
$-3+t = -2, \qquad 1-2t = -1, \qquad 1+3t = 4.$

From the first equation we have $t = 1$ . This value of $t$ also satisfies the other two equations. Hence, $(-2,-1,4) = (-3+t, 1-2t, 1+3t)$ for $t = 1$ . Therefore, $(-2,-1,4)$ is on the line.
If $(-4,3,-2)$ is on $L$ then we must have some $t \in \mathbb{R}$ such that
$-3+t = -4, \qquad 1-2t = 3, \qquad 1+3t = -2.$

From the first equation we have $t = -1$ . This value of $t$ also satisfies the other two equations. Hence, $(-4,3,-2) = (-3+t, 1-2t, 1+3t)$ for $t = -1$ . Therefore, $(-4,3,-2)$ is on the line.
If $(2,-9,16)$ is on $L$ then we must have some $t \in \mathbb{R}$ such that
$-3+t = 2, \qquad 1-2t = -9, \qquad 1+3t = 16.$

From the first equation we have $t = 5$ . This value of $t$ also satisfies the other two equations. Hence, $(2,-9,16) = (-3+t, 1-2t, 1+3t)$ for $t = 5$ . Therefore, $(2,-9,16)$ is on the line.