The equation
$$\vec r = \vec r_0 + t \vec v$$
is a parametric equation for a line. You can see $\vec r$ as a function of $t$:
$$\vec r(t) = r_0 + tv$$
and for any specific $t_0$, $\vec r(t_0)$ will be a vector pointing to a point on the line.
The whole line is the set of all points which are produced by $\vec r(t)$, i.e. the set
$$\{\vec r(t) \mid t \in \mathbb R\}.$$
Thus, it does not make sense to ask "how to solve for $t$?" as your question is phrased.
However, if you have a point $\vec p$ and ask the question "for which $t$ is $\vec r(t) = \vec p$?", that does make sense. In this case you solve the equation
$$\vec p = \vec r(t) = \vec r_0 + t \vec v$$
and this will be solvable only when $\vec p$ is on the line.
One of the main confusions in writing a line in vector form is to determine what $\vec{r}(t)=\vec{r}+t\vec{v}$ actually is and how it describes a line.
For each $t_0$, $\vec{r}(t_0)$ is a vector starting at the origin whose endpoint is on the desired line. You don't get a vector which points in the same direction as the line, but a vector ending on a line. For different $t$-values, the endpoint of the vector $\vec{r}(t)$ gives different points along the line.
On the other hand, the vector $\vec{v}=\vec{PQ}$ is parallel to the given line (even though when this vector starts at the origin, it doesn't intersect the given line).
The geometric meaning of the equation $\vec{r}(t)=\vec{r}+t\vec{v}$ is as follows: $\vec{r}$ is a vector from the origin to the line and $\vec{v}$ is a vector parallel to the desired line. $\vec{r}$ is an arrow from the origin to a point on the line. Then $t\vec{v}$ moves you from the endpoint of $\vec{r}$ in the direction of $\vec{v}$, which is parallel to the line. Since we're starting on the line after moving $\vec{r}$, this means that we're adding a vector/segment along the line.
I think of this as "giving directions", from the origin to get to a desired point on the line, walk to any point on the line (this is $\vec{r}$) and then walk in the direction of the line the desired distance (since $\vec{v}$ points in the direction of the line, $t\vec{v}$ scales the vector until it is the correct distance). The result are directions from the origin to a point on the line. Therefore, the resulting vector is not parallel to the line, but it starts at the origin and ends on the line.
Best Answer
The required vector is the position vector of the point $P$, that is the vector from the origin $O$ to $P$, i.e. : $\vec r=\vec{OP}$
The vector equation of a line is an equation that is satisfied by the vector that has its head at a point of the line. This vector is not, in general, a vector that ''lies'' on the line, unless the line passes through the origin (that is the common starting point of all vectors).
From the figure you can see that the vector $\vec r$ is such a position vector for the point $P$. For $\lambda=0$ it coincides with the vector $\vec a$(the position vector of the point $A$) and, for any other real value of $\lambda$ its head is at a point $P$ on the line that passes through $A$ and is parallel to the vector $\vec b$.
And note that $\vec r$ is the sum ( parallelogram law) of the two vectors $\vec a$ and $\lambda \vec b$.