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