Mathematics

Problem

\vec{b}

and

\vec{c}

are two non-collinear vectors such that

\vec{a} \cdot (\vec{b} + \vec{c}) = 4

and

\vec{a} \times (\vec{b} \times \vec{c}) = (x^2 - 2x + 6)\vec{b} + (\sin y)\vec{c}

, then the point

(x, y)

lies on (A)

x = 1

(B)

y = 1

(C)

y = \pi

(D)

x + y = 0