for line #vec r(lambda) = vec r_0 + lambda vec d#, the direction #vec d# of the line will be perpendicular to the normals of both planes
and so it will equal the vector cross product of their normals, calculated as the determinant of the following matrix
#det((hat i, hat j, hat k),(2,1,-1),(1,2,1))#
# = hat i (3) - hat j (3) + hat k (3) = ((3),(-3),(3))#
we can double check that by doing the cross product of #vec d# and the normals to hopefully get zero....as they are perpendicular....
#<2,1,-1>*<3,-3,3> = 6 - 3 -3 = 0# !!
and
# <1,2,1> * <3,-3,3> = 3 - 6 +3 = 0# !!
so now we have
#vec r(lambda) = vec r_0 + lambda ((3),(-3),(3))# and we just need a point #vec r_o# on the line
for each plane we can set #z=0# which reduces them to #2x +y= 3# and #x+2y = 3# which we can solve as simple linear equations. Results in #x,y = 1,1 # so the planes intersect at (1,1,0).
which yields
#vec r(lambda) = ((1),(1),(0))+ lambda ((3),(-3),(3))#
