There are two ways of evaluating a third-order determinant, but the most easiest is by;
#|(a_11quada_12quada_13), (a_21quada_22quada_23), (a_31quada_32quada_33)|=a_11|(a_22quada_23), (a_32quada_33)|-a_12|(a_21quada_23), (a_31quada_33)|+a_13|(a_21quada_22), (a_31quada_32)|#
#det=a_11(a_22a_33-a_23a_32)-a_12(a_21a_33-a_23a_31)+a_13(a_21a_32-a_22a_31)#
What needs to be remembered is that we can choose any row to break down the third-order arrangement into second-order arrangement, and it is easier when choosing a row with #0# in it because any second-order arrangement multiplied by #0# is #0#. but also we must consider its sign;
#|(a_11quada_12quada_13), (a_21quada_22quada_23), (a_31quada_32quada_33)|=|(+ - +), (- + -), (+ - +)|#
When breaking down for example into #a_11|(a_22quada_23), (a_32quada_33)|#, the #|(a_22quada_23), (a_32quada_33)|# is determined by eliminating the number that horizontally and vertically in line with #a_11#. By this way, #a_12,a_13,a_21# and #a_31# is eliminated.
So, from the question given, we break down it by using row #1# since it has #0# in it;
#|(14, -13, 0), (3, 8, -1), (-10, -2, 5)|=#
#=14|(8, -1), (-2, 5)|-(-13)|(3, -1), (-10, 5)|+0|(3, 8), (-10, -2)|#
#det=14[8(5)-(-1)(-2)]+13[3(5)-(-1)(-10)]+0[3(-2)-8(-10)]#
#det=14(38)+13(5)+0(74)#
#det=532+65#
#det=597#
