[tex]\mathbf r''(t)=\begin{bmatrix}3\\-1\\1\end{bmatrix}[/tex]
[tex]\quad\implies\mathbf r'(t)=\begin{bmatrix}3\\-1\\1\end{bmatrix}t+\mathbf C_1[/tex]
[tex]\quad\implies\mathbf r(t)=\dfrac12\begin{bmatrix}3\\-1\\1\end{bmatrix}t^2+\mathbf C_1t+\mathbf C_2[/tex]
At time [tex]t=0[/tex], the particle has position [tex]\begin{bmatrix}1\\2\\3\end{bmatrix}[/tex], so
[tex]\mathbf r(0)=\mathbf C_2=\begin{bmatrix}c_{2,1}\\c_{2,2}\\c_{2,3}\end{bmatrix}=\begin{bmatrix}1\\2\\3\end{bmatrix}[/tex]
You also know that at this point, when [tex]t=0[/tex], the particle's speed is [tex]2[/tex], which is the magnitude of the velocity at this point because the particle is traveling in one direction. This means
[tex]\left\|\mathbf r'(0)\right\|=\left\|\mathbf C_1\right\|=\left\|\begin{bmatrix}c_{1,1}\\c_{1,2}\\c_{1,3}\end{bmatrix}\right\|=\sqrt{{c_{1,1}}^2+{c_{1,2}}^2+{c_{1,3}}^2}=2[/tex]
If [tex]T[/tex] is the time at which the particle arrives at the point [tex](4,1,4)[/tex], i.e. [tex]\mathbf r(T)=\begin{bmatrix}4\\1\\4\end{bmatrix}[/tex], then you have the system
[tex]\begin{cases}\dfrac32T^2+c_{1,1}T+1=4\\\\-\dfrac12T^2+c_{1,2}T+2=1\\\\\dfrac12T^2+c_{1,3}T+3=4\end{cases}[/tex]
Eliminating the terms with [tex]T^2[/tex], you're left with the system
[tex]\begin{cases}c_{1,1}+3c_{1,2}=0\\\\c_{1,1}-3c_{1,3}=0\\\\c_{1,2}+c_{1,3}=0\end{cases}[/tex]
so you're left with solving this system with the constraint that the sum of these constants' squares is [tex]2^2=4[/tex].
However, there are two solutions to this, with [tex]\mathbf C_1=\pm\dfrac1{\sqrt{11}}\begin{bmatrix}6\\-2\\2\end{bmatrix}[/tex]
Obviously, we require that [tex]T>0[/tex], so we need to check if either option forces using an invalid value of [tex]T[/tex]. This amounts to finding [tex]T[/tex] such that [tex]\mathbf r(T)=(4,1,4)[/tex]. Indeed, this has two solutions, [tex]T=\dfrac{-2\sqrt{11}\pm\sqrt{286}}{11}[/tex], or [tex]T\approx-2.14[/tex] and [tex]T\approx0.934[/tex]. We ignore the first.
Plugging in this value of [tex]T[/tex] into the system [tex]\mathbf r(T)=(4,1,4)[/tex], we find that [tex]\mathbf C_1=\dfrac1{\sqrt{11}}\begin{bmatrix}6\\-2\\2\end{bmatrix}[/tex]
Therefore the equation for the position vector is
[tex]\mathbf r(t)=\begin{bmatrix}\dfrac32t^2+\dfrac6{\sqrt{11}}t+1\\\\-\dfrac12t^2-\dfrac2{\sqrt{11}}t+2\\\\\dfrac12t^2+\dfrac2{\sqrt{11}}t+3\end{bmatrix}[/tex]
