Divide the face of the clock into three parts with two lines so that the sum of the numbers in the three parts are equal.

Let P be a point between 10 and 11

Let Q be a point between  2and 3

Draw the chord PQ.

Let R be a point between 8 and 9 and S be a point between  4 and 5, Join the chord RS

Now the face of the clock has 3 regions.  The sector bounded by the chord PQ and its arch ,  the region PQSR   ,  and the region enclosed between the chord RS and its arch  containing numbers respectively: (2,1,,12,and 11) , (3, 4, 10 and  9 )  and (8, 7, 6 and 5)  in each regions so divided. The sum in each part or region is  26.

If the the clock has 12 not marked , then it could be treated as  zero and then clock face contains 1 to 11 and a  no number or zero. Under this situation,  [1, 0 (or no number),  11  , 10] , [3,  2,  9,  8] and [7,  6,  5   4] are the  numbers in the 3 regions separated by two chord PQ ( P in between  9 and 10 ,  Q is in between 1 and 2)  and RS (R is in between 7 and 8,     and S between  3 and 4). In this case the total of numbers in each region is 22.