If the 3 digit number formed by deleting the colored digits is divisible by 7, we can obtain a multiple of 7 using the 1st option. If else, we we reorder the digits such that the 3 uncolored digits are in the last 3 positions, and the colored digits are arranged in way such that the the first 3 digits and the last 3 digits have the same remainder by 7. This works because the digits 1,2, and 4 can be arranged to produce every remainder except for 0 mod 7, and since $1000\equiv -1$ $(mod $ $7)$.