The notation in the problem statement seems misleading to me.
We are given symbols $P_1,$ $P_2,$ $V_1,$ $V_2,$ $v_1,$ and $v_2,$
seemingly describing objects $1$ and $2$ in a completely symmetric fashion,
whereas the problem statement actually is not symmetric.
The problem statement implies that $P_1,$ $P_2,$ $V_1,$ $v_1,$ and $v_2$
are all known, with $v_1 = \lVert V_1\rVert.$
The unknown entity is $V_2$; we have the constraint that
$\lVert V_2\rVert = v_2,$ but as long as that constraint is satisfied,
we can choose the direction of $V_2$ freely.
The mathematics of finding a suitable direction for $V_2$ were thoroughly hashed out in the answers to Intersection of two moving objects.
There is an algebraic treatment
which concludes with an equation that the angle $\phi$ (the direction of vector $V_2$) must meet in order to make Object $2$ meet Object $1.$
There are at most two solutions of the equation, but in some cases a solution of the equation gives a value of $\phi$ that causes Object $2$ to move away from Object $1$ (so they will never meet), while in other cases there are two values of $\phi$ that would cause the objects to meet, although only one of those values can be the one that makes them meet in the minimum possible time.
I find a geometric approach gives me more insight on the problem;
in particular, once the geometry of the problem has been set up
in this way
(see this answer
or this answer)
it immediately gives us the equation
$$
v_2 \cos \phi = -v_1 \cos\theta,
$$
where $\phi$ is the angle between $V_2$ and the line $P_2P_1$
and $\theta$ is the angle between $V_1$ and the line $P_2P_1,$
measuring both angles counterclockwise.
(The algebraic treatment comes to an equivalent conclusion.)
Another way to say this is:
In order for the two objects to meet, the component of $V_2$ perpendicular to the line between the objects must exactly cancel the component of $V_1$ perpendicular to the line between the objects.
This condition is necessary but not sufficient to solve the problem.
A condition that is both necessary and sufficient is:
The vector $V_2 - V_1$ must point in the direction from $P_2$ to $P_1.$
Geometrically, we can construct representations of all possible vectors of the form $V_2 - V_1$ (given that $V_1$ is fixed and the magnitude but not the direction of $V_2$ is fixed)
by drawing the vector $-V_1$ and then drawing a circle of radius $v_2$
around the tip of $V_1.$
We can then represent $V_2 - V_1$ by placing its tail at the tail of
$-V_1$ and its head at any point on the circle; the direction from the center of the circle to the chosen point is the direction of $V_2.$
The diagram in this answer
shows the case where $v_1 > v_2$ but where it is still possible to make the objects meet.
In this case the circle intersects the line $P_1P_2$ in two places,
giving two possible choices of a vector $V_2$ that will cause the objects to meet.
The choice that causes the objects to meet in the minimum possible time is the choice for which the magnitude of $V_2 - V_1$ is larger.
(In the diagram in the linked answer, that vector is $CG.$)
When $v_1 > v_2,$ it can also happen that there is no solution to the problem; object $2$ can only "catch" object $1$ if object $1$ is initially reducing its distance to object $2$ quickly enough.
When $v_2 \geq v_1,$ there is always a solution but only one solution
for $V_2$ that allows object $2$ to "catch" object $1,$
so of course that is also the minimum-time solution.
The solution is shown graphically in this answer.
