The quadrilateral formed by the points $A(1,2,5), B(-1,6,1), C(3,4,-3)$ and $D(5,0,1)$ is a

A line with direction cosines proportional to $2,1,2$ meets the line $L_1$ passing through $(0,-1,0)$ with direction ratios $1,1,1$ at $A(x, y, z)$ and another line $L_2$ at $B(1,1,1)$ then $x+y+z=$

If a plane $\pi$ passes through the point $(-1,6,2)$ is perpendicular to the planes $x+2 y+2 z-5=0$ and $3 x+3 y+2 z-8=0$, then, the perpendicular distance from the point $(1,-1,1)$ to the plane $\pi$ is

If $f: \mathbf{R} \rightarrow \mathbf{R}$ and $g: \mathbf{R} \rightarrow \mathbf{R}$ be defined by $f(x)=\left\{\begin{array}{cc}x+2, & x>0 \\ 2-x, & x \leq 0\end{array}\right.$ and $g(x)=\left\{\begin{array}{cc}x^2-2 x-2, & 1 \leq x<2 \\ x-7 & x \geq 2 \\ x+5, & x<1\end{array}\right.$ then $\lim _{x \rightarrow 0} g \circ f(x)$