Given that #vec a + 2vec b + 3vec c = vec O#
Now, Taking cross product or vector product with #vec a # on both sides:-
#(vec a + 2vec b + 3vec c ) xx vec a = vec O xx vec a#
Since cross product is distributive and cross product with #vec O# OR null vector is #vec O# itself,
#=> vec a xx vec a + 2(vec b xx vec a) + 3(vec c xx vec a) = vec O#
#because # the cross product of a vector with itself is null vector or #vec O# #therefore# #vec a xx vec a = vec O#
#=> 2(vec b xx vec a) + 3(vec c xx vec a) = vec O# ------------------- 1.
Similarly, taking cross product with #vec b and vecc# on both sides:-
#(vec a xx vec b) + 3(vec c xx vec b) = vec O# ------------ 2.
#(vec a xx vec c) + 2(vec b xx vec c) = vec O# ------------ 3.
Adding 1., 2. and 3.
#2(vec b xx vec a) + 3(vec c xx vec a) + (vec a xx vec b) + 3(vec c xx vec b) + (vec a xx vec c) + 2(vec b xx vec c) = vec O#
Now for any two vectors #vec x# and #vec y#, #color(red)(vec x xx vec y = - vec y xx vec x)#.
Also, sum of any vector with null vector is the vector itself.
#=> (vec a xx vec b) - 2(vec a xx vec b) + 2(vec b xx vec c) - 3(vec b xx vec c) + 3(vec c xx vec a) - (vec c xx vec a) = vec O#
#=> -(vec a xx vec b) - (vec b xx vec c) - (vec c xx vec a) + 3(vec c xx vec a) = vec O#
#=> -(vec a xx vec b) - (vec b xx vec c) - (vec c xx vec a) = - 3(vec c xx vec a) #
Taking product on both sides with #-1#,
#=> vec a xx vec b + vec b xx vec c + vec c xx vec a = 3(vec c xx vec a)#
