if
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#vec n# is the normal vector to the plane #pi#,
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#r_o# is a specific point lying on #pi#,
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#vec r# is any other point on #pi#
then #(\vec r - vec r_o) # is a line lying along #pi#
....and so we can say using the scalar dot product that
#(\vec r - vec r_o) * vec n = 0#, which is the vector equation of the plane #pi#
we can re-arrange that to #\vec r * vec n - vec r_o * vec n = 0#
and, in Cartesian with #vec r = ((x),(y),(z))#, we can match it up to your plane as follows using the scalar dot product
#((x),(y),(z)) * ((2),(-3),(1)) - 6 = 0#
for a parametric form, we will need a point #vec p_o# on plane #pi#, and 2 non-parallel vectors, #vec u# and #vec v#, that run along #pi#. We can then identify any point on #pi# as
#vec r = vec p_0 + s vec u + t vec v# where s and t are the paremeters.
generating a #vec p_o # is simple. we just take #2x -3y + z - 6 = 0# and set #x = y = 0# so that #vec p_o = ((0),(0),(6))#
next we want #vec u# and # vec v# to be orthogonal to #vec n#
Again using the scalar dot product, that means #vec u * vec n = vec v * vec n = 0#
So for #vec u# we have #2 u_1 - 3 u_2 + u_3 = 0#
We can just arbitrarily choose #u_1 = 1, u_2 = 0 implies u_3 = -2# giving #vec u = ((1),(0),(-2))#
For #vec v# we have #2 v_1 - 3 v_2 + v_3 = 0# but we must ensure that #vec v# is not parallel to #vec u#
So this time we can just arbitrarily choose #v_1 = 0, v_2 = 1 implies v_3 = 3# giving #vec v = ((0),(1),(3))#
Now we can check our work!! We would expect the result #vec u times vec v = vec n# as #vec u# and #vec v# lie in #pi# and #vec n# is it's normal vector
#vec u times vec v = det ((hat x, hat y, hat z),(1,0,-2),(0,1,3))#
#=2 hat x - 3 hat y + 1 hat z# BINGO!!!!
one form of parameterised plane #pi# is therefore
#pi: vec r = ((0),(0),(6)) + s ((1),(0),(-2)) + t ((0),(1),(3)) #
