If $\alpha$ and $\beta$ are the angles made by the normals drawn from the origin to the lines $x+y+\sqrt{2}=0$ and $x-\sqrt{3}y-2=0$ with the positive direction of the $X$-axis respectively,measured in the anti-clockwise direction,then $\alpha+\beta=$

The coordinates of the foot of the perpendicular from the point $(1, 2)$ on the line $x - 3y + 7 = 0$ are

Let $\alpha$ and $\beta$ be integers satisfying $0 < \beta < \alpha$. Let $P(\alpha, \beta)$ be a point. Let $Q$ be the reflection of $P$ in the line $y = x$,$R$ be the reflection of $Q$ in the $y$-axis,$S$ be the reflection of $R$ in the $x$-axis,and $T$ be the reflection of $S$ in the $y$-axis. If the area of the convex pentagon $PQRST$ is $187 \ sq. \ units$,then the value of $\alpha + \beta^2$ is:

The unit normal vector to the line joining $i - j$ and $2i + 3j$ and pointing towards the origin is

The coordinates of the foot of the perpendicular from $(0,0)$ upon the line $x+y=2$ are

Let $A \left(\frac{3}{\sqrt{a}}, \sqrt{a}\right)$ with $a > 0$ be a fixed point in the $xy$-plane. The image of $A$ in the $y$-axis is $B$,and the image of $B$ in the $x$-axis is $C$. If $D(3 \cos \theta, a \sin \theta)$ is a point in the fourth quadrant such that the maximum area of $\triangle ACD$ is $12$ square units,then $a$ is equal to: