The condition for \(\triangle A B C\) to have a right angle at \(B\) is that lines \(\overrightarrow{A B}\) and \(\overleftrightarrow{B C}\) are perpendicular, which holds when and only when the product of their slopes is \(-1\), that is \(\left(\frac{3}{8}\right)[-2 /(1+k)]=-1\). This is equivalent to \(6=8(1+k), \quad\) or \(8 k=-2\), or \(k=-\frac{1}{4}\).