Let $\vec b$ and $\vec c$ be non-collinear vectors. If $\vec a$ is a vector such that $\vec a.(\vec b+\vec c)=4$ and $\vec a×(\vec b×\vec c)=(x^2-2x+6)\vec b+\sin y.\vec c$, then (x, y) lies on the line.

x = 1

According to the definition

$\vec a×(\vec b×\vec c)=(\vec a.\vec c)\vec b-(\vec a.\vec b)\vec c$

given in the question

$(\vec a.\vec c)\vec b-(\vec a×\vec b)\vec c=(x^2-2x+6)\vec b+(\sin y)\vec c$

On comparing

$\vec a.\vec c=x^2-2x+6$

$\vec a.\vec b=-\sin y$

$\vec a.(\vec b+\vec c)=4⇒\vec a.\vec b+\vec a.\vec c=4$

$⇒-\sin y+x^2-2x+6=4$

$⇒(x-1)^2+1=\sin y$

For this to be minimum (x - 1)² = 0

⇒ x = 1

Minimum value of sin y = 1 $⇒y=\frac{\pi}{2}$

$⇒(x-1)^2+1=\sin y$ is possible when x = 1 & $y=\frac{\pi}{2}$