Let $A(4,3,5), B(1,-2,1), C(3,2,1)$ be the vertices of a $\triangle A B C$. If the internal bisector of $\angle B A C$ meet the side $B C$ at $D$, then $C D=$

A line $L$ has intercepts $a$ and $b$ on the coordinate axes. When the coordinate axes are rotated through an angle $\alpha$ and keeping the origin fixed, the same line $L$ has intercepts $p$ and $q$ on the new axes. Then,

Two lines $L_1$ and $L_2$ passing through the point $P(1,2)$ cut the line $x+y=4$ at a distance of $\frac{\sqrt{6}}{3}$ units from $P$. Then, the angles made by $L_1, L_2$ with positive $X$-axis are

A pair of straight lines drawn though the origin forms. an isosceles triangle right angled at the origin with the line $2 x+3 y=6$. The area (in sq units) of the triangle, so formed is