Prove some properties of a given point and line

Let $P = (1,2,3), \ A = (1,-2,2), \ Q = (3,3,1)$ and consider the line $L(P;A)$ . Let $X(t) = P + tA$ be an arbitrary point on this line.

Compute $\lVert Q - X(t) \rVert^2$ .
Prove there is exactly one point $X(t_0)$ such that $\lVert Q - X(t) \rVert$ is minimal, and compute the minimum distance.
Prove that $Q - X(t_0)$ is orthogonal to $A$ .

We compute,
$\begin{align*} X(t) = P + tA && \implies && X(t) &= (1+t, 2-2t, 3+2t) \\ && \implies && Q-X(t) &= (2-t, 1+2t, -2-2t) \\ && \implies && \lVert Q - X(t) \rVert^2 &= \big( (2-t, 1+2t, -2-2t) \cdot (2-t, 1+2t, -2-2t) \big) \\ &&&&&= 4-4t + t^2 + 1 + 4t + 4t^2 + 4 + 8t + 4t^2 \\ &&&&&= 9+8t + 9t^2. \end{align*}$
Proof. From part (a) we know the distance function is $d(t) = 9 + 8t + 9t^2$ . Therefore, the derivative is $d'(t) = 8 + 18t$ . This derivative is less than 0 for $t < -\frac{4}{9}$ and greater than 0 for $t > -\frac{4}{9}$ . Therefore, $\lVert Q - X(t) \rVert$ has a minimum at $t = -\frac{4}{9}$ . The distance at this point is
$\left( 9 + 8 \left( -\frac{4}{9} \right) + 9 \left( -\frac{4}{9} \right)^2 \right)^{\frac{1}{2}} = \frac{\sqrt{65}}{3}. \qquad \blacksquare$
Proof. First, we have
$\begin{align*} Q - X(t_0) &= (3,3,1) - (P + t_0 A) \\ &= (3,3,1) - \left( (1,2,3) - \frac{4}{9} (1,-2,2) \right) \\ &= (3,3,1) - \left( \frac{5}{9}, \frac{26}{9}, \frac{19}{9} \right) \\ &= \left( \frac{22}{9}, \frac{1}{9}, -\frac{10}{9} \right). \end{align*}$

Therefore,

$\begin{align*} A \cdot (Q - X(t_0)) &= (1,-2,2) \cdot \left( \frac{22}{9}, \frac{1}{9}, -\frac{10}{9} \right) \\ &= \frac{22}{9} - \frac{2}{9} - \frac{20}{9} \\ &= 0. \end{align*}$

Hence, $A$ and $Q-X(t_0)$ are orthogonal $. \qquad \blacksquare$