Given point X on line L, compute distance between Q and X (as a function of t). Where is my error?

Question

Let $P$ be the point $(1,-1,3,1)$, $Q$ be the point $(1,1,-1,2)$ and $A$ be the vector $(1,-3,2,1)$.

$(a)$ Given point $X$ on line $L$, compute the distance between $Q$ y $X$ (as a function of t).

Distance between point $X$ and $Q$.

$$L=(1,-1,3,1)+t(1,-3,2,1)$$ $$X=(1+t,-1-3t,3+2t,1+t)$$ $$\overline{QX}=(t,-2-3t,4+2t,-1+t)$$ $$\lVert\overline{QX}\rVert=\sqrt{t²+(-2-3t)²+(2t+4)²+(t-1)²}=\sqrt{16t²+26t+21}$$

Minimum distance in my computation is when $t=-1\implies\sqrt{11}$ but the correct minimum is $\sqrt{\frac{146}{15}}$.

Answer 1

To obtain the distance between $Q$ and $X$, you should have $QX \cdot A = 0$ as the vector $QX$ must be perpendicular to the line to find the minimum distance. This implies that $t-3(-2-3t)+2(4+2t)-1+t = 0$, hence $t=-13/15$.