If the circles $$(x+1)^2+(y+2)^2=r^2$$ and $$x^2+y^2-4 x-4 y+4=0$$ intersect at exactly two distinct points, then

Let $$\overrightarrow{\mathrm{a}}=\mathrm{a}_1 \hat{i}+\mathrm{a}_2 \hat{j}+\mathrm{a}_3 \hat{k}$$ and $$\overrightarrow{\mathrm{b}}=\mathrm{b}_1 \hat{i}+\mathrm{b}_2 \hat{j}+\mathrm{b}_3 \hat{k}$$ be two vectors such that $$|\overrightarrow{\mathrm{a}}|=1, \vec{a} \cdot \vec{b}=2$$ and $$|\vec{b}|=4$$. If $$\vec{c}=2(\vec{a} \times \vec{b})-3 \vec{b}$$, then the angle between $$\vec{b}$$ and $$\vec{c}$$ is equal to:

The maximum area of a triangle whose one vertex is at $$(0,0)$$ and the other two vertices lie on the curve $$y=-2 x^2+54$$ at points $$(x, y)$$ and $$(-x, y)$$, where $$y>0$$, is :

If $$f(x)=\left|\begin{array}{ccc} 2 \cos ^4 x & 2 \sin ^4 x & 3+\sin ^2 2 x \\ 3+2 \cos ^4 x & 2 \sin ^4 x & \sin ^2 2 x \\ 2 \cos ^4 x & 3+2 \sin ^4 x & \sin ^2 2 x \end{array}\right|,$$ then $$\frac{1}{5} f^{\prime}(0)=$$ is equal to :