Let three vectors ,$$\overrightarrow{\mathrm{a}}=\alpha \hat{i}+4 \hat{j}+2 \hat{k}, \overrightarrow{\mathrm{b}}=5 \hat{i}+3 \hat{j}+4 \hat{k}, \overrightarrow{\mathrm{c}}=x \hat{i}+y \hat{j}+z \hat{k}$$ form a triangle such that $$\vec{c}=\vec{a}-\vec{b}$$ and the area of the triangle is $$5 \sqrt{6}$$. If $$\alpha$$ is a positive real number, then $$|\vec{c}|^2$$ is equal to:

Let $$f(x)=a x^3+b x^2+c x+41$$ be such that $$f(1)=40, f^{\prime}(1)=2$$ and $$f^{\prime \prime}(1)=4$$. Then $$a^2+b^2+c^2$$ is equal to:

The parabola $$y^2=4 x$$ divides the area of the circle $$x^2+y^2=5$$ in two parts. The area of the smaller part is equal to :

The solution of the differential equation $$(x^2+y^2) \mathrm{d} x-5 x y \mathrm{~d} y=0, y(1)=0$$, is :