7. Given $\boldsymbol{A} \boldsymbol{B}=(k, 1), \boldsymbol{A} \boldsymbol{C}=(2,3)$. Then the value of $k$ that makes $\triangle A B C$ a right triangle is ( ).
(A) $\frac{3}{2}$
(B) $1-\sqrt{2}$
(C) $1-\sqrt{3}$
(D) $-\sqrt{5}$
7.C.

If $\angle B A C=90^{\circ}$, then $A B \cdot A C=2 k+3=0$.
Solving for $k$ gives $k=-\frac{3}{2}$.
If $\angle A B C=90^{\circ}$, then
$$
C B \cdot A B=(A B-A C) \cdot A B=k^{2}-2 k-2=0 \text {. }
$$

Solving for $k$ gives $k=1 \pm \sqrt{3}$.
If $\angle A C B=90^{\circ}$, then
$$
C B \cdot A C=(A B-A C) \cdot A C=2 k-10=0 \text {. }
$$

Solving for $k$ gives $k=5$.