A plane meets the coordinate axes at $A, B, C$ respectively such that the centroid of the $\triangle A B C$ is $(2,3,5)$. Then, the equation of that plane is

$L_1$ is a line passing through the points with position vectors $\hat{\mathbf{i}}-2 \hat{\mathbf{j}}-\hat{\mathbf{k}}$ and $4 \hat{\mathbf{i}}-3 \hat{\mathbf{k}} . L_2$ is a line passing through the points with position vectors $\hat{\mathbf{i}}+2 \hat{\mathbf{j}}-\hat{\mathbf{k}}$ and $2 \hat{\mathbf{i}}-4 \hat{\mathbf{j}}-5 \hat{\mathbf{k}}$. Then the distance between $L_1$ and $L_2$ is

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 $x+y+z=$