The plane equation can be put in a formulation more useful
#Pi->x+3y+z-6 equiv << p -p_0, vec n >> = 0#
with
#p = {x,y,z}# a generic plane point
#p_0 = {x_0,y_0,y_0}# given plane point
#vec n = {1,3,1}# normal plane vector.
The determination of #p_0# is straightforward
#(x-x_0)+3(y-y_0)+(z-z_0) =x+3y+z-(x_0+3y_0+z_0) = x+3y+z-6#
so
#x_0+3y_0+z_0=6#. Fixing #x_0,y_0# we obtain #z_0=6# so
#p_0 = {0,0,6}#
The three points limiting the plane area contained in the first quadrant are:
#p_1={6,0,0}#
#p_2={0,2,0}#
#p_3={0,0,6}#
Those points are obtained fixing to zero two coordinates into the plane equation, and computing the third.
Considering #p_1,p_2,p_3# as the vertices of the piramid whose cusp vertice is the origin of coordinates, knowing the distance between this vertice and the basis plane, we will obtain the pyramid height and consequently we will be able to compute its volume.
The height or distance between the origin and the plane is obtained as follows.
Given a plane # << p -p_0, vec n >> = 0# and a point #q# the distance between #q# and the plane is given by
#norm(q-p_i)# where #p_i# is given by the solution of
# << p_i -p_0, vec n >> = 0# and #p_i=q+lambda vec n#
substituting we have
#<< q+lambda vec n-p_0,vec n >> = << q,vec n >> + lambda << vec n, vec n >> - << p_0, vec n >> = 0# for
#lambda =( << p_0, vec n >> - << q,vec n >> )/<< vec n, vec n >># but here #q = {0,0,0}# then
#lambda = << p_0, vec n >> /<< vec n, vec n >># so #p_i# is
#p_i = q+ << p_0, vec n >> /<< vec n, vec n >> vec n# and
#d = norm(p_i-q) = norm(<< p_0, vec n >> /<< vec n, vec n >> vec n) = abs(<< p_0, vec n >>/norm(vec n)) = 6/sqrt(11)#.
The basis area is given by
#A=1/2norm((p_2-p_1) xx (p_3-p_1)) = 6 sqrt(11)#
and finally the pyramid volume is given by
#V = 1/3 cdot d cdot A = 12#
