The first mosquito trajectory is modelled by the $3D$ line
$r_A(t) = r_1 + t V_1 $
And the second mosquito trajectory is modelled by
$ r_B(t) = r_2 + t V_2 $
Note that $t$ is the same variable in both equations. Also note that $r_1, r_2, V_1, V_2 $ are known quantities.
The "missile" or "spray" is modelled by
$ r_C(t) = r_3 + t V_3 $
where $r_3$ and $V_3$ are unknown, however it is known that $|V_3| = s $ (i.e. the speed of the interceptor (the missile) is known). And the objective is to find $r_3$ and $V_3$ such that the distance travelled by the missile after hitting $r_A$ till it hits $r_B$ is minimum.
Equations:
Let $t_1$ and $t_2$ be the interception ("hitting") times for $r_A$ and $r_B$ then
$ r_1 + t_1 V_1 = r_3 + t_1 V_3 \tag{1} $
$ r_2 + t_2 V_2 = r_3 + t_2 V_3 \tag{2}$
Hence,
$ r_2 + t_2 V_2 - r_1 - t_1 V_1 = (t_2 - t_1) V_3 \tag{3} $
And immediately, we have a condition on $t_1$ and $t_2$ that follows from the fact that $V_3^T V_3 = s^2 $
Namely,
$ (r_2 + t_2 V_2 - r_1 - t_1 V_1)^T (r_2 + t_2 V_2 - r_1 - t_1 V_1) = (t_2 - t_1)^2 s^2 \tag{4} $
This is an equation of a conic in the $t_1 t_2$ plane. It could be an ellipse, a hyperbola, or a parabola.
We're interested in minimizing $t_2 - t_1$. The contours of this function (of two variables) are straight lines that make an angle of $45^\circ$ with the $t_1$ axis, and the value of $t_2 - t_1$ decreases as the contour line moves down and right.
Once the tangency point is determined, then $t_1$ and $t_2$ are known, and everything else follows from it.
Applying this method to the numerical values given in this problem, we have:
$ r_1 = (-2, 2, 3) , V_1 = (6, -4, 5) $
$ r_2 = (3,5,1) , V_2 = (-7, -3, 4) $
$s = \sqrt{25.25}$
The tangent point to the ellipse is
$(t_1, t_2) = (0.152823, 0.556005)$
So
$ V_3 = (0.473803, 4.819857, -1.33959) $
and
$ r_3 = (-1.15547, 0.652123, 3.968835) $
The minimum distance is
$ d = (t_2 - t_1) s = (0.556005-0.152823) (\sqrt{25.25} ) = 2.025963 \text{ units}$
The figure below shows the ellipse on which $t_1$ ($x$ coordinate) and $t_2$($y$ coordinate) lie. We're only concerned with the part of this ellipse that satisfies $t_2 \gt t_1 \ge 0 $. The green line is the tangent line to the ellipse having a slope of $1$.
The critical point (the tangent point) is the point colored in red. It is very close to the vertex of the ellipse along the major axis of the ellipse because of the numerical values of vectors $V_1$ and $V_2$.
If we change the numerical values as follows:
$ r_1 = (-2, 2, 3) , V_1 = (6, -4, 5) $
$ r_2 = (3,5,1) , V_2 = (-4, -2, 1) $
$s = \sqrt{25.25}$
Then the resulting relation between $t_1$ and $t_2$ will be a hyperbola, which is shown below
In this case, the resulting values of $t_1$ and $t_2$ corresponding to the tangent point (the red dot in the graph) are:
$t_1 = 0.137357 $
$ t_2 = 0.747268 $
And this gives
$ V_3 = (1.945835, 3.36917, -3.18) $
$ r_3 = (-1.44313, 0.987792, 4.123582) $
The minimum possible distance between the two interceptions is
$ d = (t_2 - t_1) s = (0.747268 - 0.137357) \sqrt{25.25} = 3.064763 \text{ units}$
Noteworthy:
The problem of finding the minimum of the function $(t_2 - t_1)$ subject to equation $(4)$ can be approached using Lagrange multiplier method. The objective function is defined as
$ f(t_1, t_2, \lambda) = t_2 - t_1 + \lambda g(t_1, t_2) \tag{5}$
where
$ g(t_1, t_2) = (r_2 + t_2 V_2 - r_1 - t_1 V_1)^T (r_2 + t_2 V_2 - r_1 - t_1 V_1) - (t_2 - t_1)^2 s^2 \tag{6}$
The function $g$ is quadratic in $t_1$ and $t_2$, and can be written compactly as
$ g(t_1, t_2) = x^T A x + x^T B + C $
where
$ x = \begin{bmatrix} t_1 \\ t_2 \end{bmatrix} $
$ A = \begin{bmatrix} V_1^T V_1 - s^2 && - V_1^T V_2 + s^2 \\ - V_1^T V_2 + s^2 && V_2^T V_2 - s^2 \end{bmatrix} $
$ B = \begin{bmatrix} -2 V_1^T (r_2 - r_1) \\ 2 V_2^T (r_2 - r_1) \end{bmatrix}$
$ C = (r_2 - r_1)^T (r_2 - r_1) $
Taking the gradient of the function defined in $(5)$, we get
$ \nabla_x f = N + \lambda \bigg( 2 A x + B \bigg) $
where $N = \begin{bmatrix} -1 \\ 1 \end{bmatrix} $
Setting the gradient equal to the zero vector, gives us
$ N = - \lambda \bigg( 2 A x + B \bigg) $
In other words, $N$ and $(2 A x + B)$ are along the same direction (parallel or opposite). This can only happen if
$ N \times (2 A x + B) = \mathbf{0} $
where the 2-vectors have been appended with a zero third component, so that the cross product is well-defined.
This cross multiplication is identical to the following matrix multiplication:
$ S_N (2 A x + B) = \mathbf{0} $
where
$ S_N = \begin{bmatrix} 0 && - N_z && N_y \\ N_z && 0 && -N_x \\ - N_y && N_x && 0 \end{bmatrix} = \begin{bmatrix} 0 && 0 && 1 \\ 0 && 0 && 1 \\ -1 && -1 && 0 \end{bmatrix} $
Note that the first two components of $ S_N (2 A x + B) $ will automatically be zero from the structure of $S_N$. Therefore, we only need to concentrate on the third components. It is straight forward to see that the third row of the above matrix-vector equation reads
$ 2 \bigg( (A_{11} + A_{21} ) t_1 + (A_{12} + A_{22} ) t_2 \bigg) + B_1 + B_2 = 0 \tag{7}$
Equation $(7)$ gives an additional relation (which is linear) between the variables $t_1$ and $t_2$.
All we have to do is solve the system of equations given by the linear equation $(7)$ and the quadratic equation $g(t_1, t_2) = 0 $.
Once we do that, we'll get $2$ solutions (or only one solution in case $g(t_1, t_2) = 0 $ represents a parabola).
Next, we have to take each of these two solutions, and validate it, that is , make sure that $t_2 \gt t_1 \ge 0 $. This validation process will result in the required solution.
Finally, once we have $t_1$ and $t_2$ we can compute $V_3$ from equation $(3)$, and then we can compute $r_3$ from $(1)$ or $(2)$.
