A vector $$\overrightarrow a = \left( {x,\,y,\,z} \right)$$    of length $$2\sqrt 3 $$  which makes equal angles with the vectors $$\overrightarrow b = \left( {y,\, - 2z,\,3x} \right)$$    and $$\overrightarrow c = \left( {2z,\,3x,\, - y} \right)$$    and is perpendicular to $$\overrightarrow d = \left( {1,\, - 1,\,2} \right)$$    and makes an obtuse angle with $$y$$-axis is :

Solution :
Since, $$\overrightarrow a $$ is $$ \bot $$ to $$\overrightarrow d ,$$ so $$x - y + 2z = 0......\left( 1 \right)$$
Moreover, $$\left| {\overrightarrow b } \right| = \left| {\overrightarrow c } \right|,$$   so $$\overrightarrow a .\overrightarrow b = \overrightarrow a .\overrightarrow c $$
as $$\overrightarrow a $$ makes equal angles with $$\overrightarrow b $$ and $$\overrightarrow c .$$ Thus
$$\eqalign{ & xy - 2yz + 3xz = 2xz + 3xy - yz \cr & \Rightarrow xz - 2xy - yz = 0......\left( 2 \right) \cr & {\text{Also, }}{x^2} + {y^2} + {z^2} = 12......\left( 3 \right) \cr & {\text{and }}y < 0 \cr} $$
Put the value of $$y$$ from equation $$\left( 1 \right)$$ in equation $$\left( 2 \right),$$
we get, $${x^2} + 2xz + {z^2} = 0\,;$$
So, $$x = - z{\text{ and }}y = z$$
Again put these values in equation $$\left( 3 \right),$$
we get $${z^2} = 4 \Rightarrow z = \pm 2$$
But $$y < 0$$  and $$y = z$$
Hence, $$z = - 2 = y{\text{ and }}x = 2$$