Let ABC be an equilateral triangle with orthocenter at the origin and the side BC on the line $x+2 \sqrt{2} y=4$. If the co-ordinates of the vertex A are $(\alpha, \beta)$, then the greatest integer less than or equal to $|\alpha+\sqrt{2} \beta|$ is

Let the angles made with the positive $x$-axis by two straight lines drawn from the point $\mathrm{P}(2,3)$ and meeting the line $x+y=6$ at a distance $\sqrt{\frac{2}{3}}$ from the point P be $\theta_1$ and $\theta_2$. Then the value of $\left(\theta_1+\theta_2\right)$ is:

Let $A(1,0), B(2,-1)$ and $C\left(\frac{7}{3}, \frac{4}{3}\right)$ be three points. If the equation of the bisector of the angle ABC is $\alpha x+\beta y=5$, then the value of $\alpha^2+\beta^2$ is

Let $\mathrm{A}(1,2)$ and $\mathrm{C}(-3,-6)$ be two diagonally opposite vertices of a rhombus, whose sides AD and BC are parallel to the line $7 x-y=14$. If $\mathrm{B}(\alpha, \beta)$ and $\mathrm{D}(\gamma, \delta)$ are the other two vertices, then $|\alpha+\beta+\gamma+\delta|$ is equal to :