B

$\because$二次函数过$(-10$，$y_{1})$，$(2$，$y_{2})$，$(-1$，$y_{3})$，$(-5$，$y_{1})$，
$\therefore$对称轴为直线$x=-\frac{15}{2}$，图象开口向下，
$\therefore$离对称轴$x=-\frac{15}{2}$越远，函数值越小，
$\because |-10+\frac{15}{2}|＜|-1+\frac{15}{2}|＜|2+\frac{15}{2}|$，
$\therefore (-10$，$y_{1})$，$B(2$，$y_{2})$，$(-1$，$y_{3})$三点中，$(-10$，$y_{1})$离对称轴最近，$B(2$，$y_{2})$离对称轴最远.
综上所述：$y_{2} \lt y_{3} \lt y_{1}$.
故选：$B$.