If Positive Electric Charges of Q and Q Coulombs Are $( a , b , c )$

If positive electric charges of Q and q coulombs are situated at positions $( a , b , c )$ and $( x , y , z )$ , respectively, then the force of repulsion they experience is given by $F = K \frac { Q q } { ( x - a ) ^ { 2 } + ( y - b ) ^ { 2 } + ( z - c ) ^ { 2 } }$
Where $K \approx 9 \times 10 ^ { 9 }$ , F is given in newtons, and all positions are measured in meters. Assume that a charge of 6 coulombs is situated at the origin and that a second charge of 12 coulombs is situated $( 6,6,4 )$ and moving in the y direction at 1.5 meters per second. How fast is the electrostatic force it experiences decreasing

Round your answer to the nearest trillion.

A) $1.506198347 \times 10 ^ { 18 } \frac { \mathrm { N } } { \mathrm { s } }$
B) $4.506198347 \times 10 ^ { 8 } \frac { \mathrm { N } } { \mathrm { s } }$
C) $1.506198347 \times 10 ^ { 9 } \frac { \mathrm { N } } { \mathrm { s } }$
D) $3.506198347 \times 10 ^ { 9 } \frac { \mathrm { N } } { \mathrm { s } }$
E) $2.506198347 \times 10 ^ { 10 } \frac { \mathrm { N } } { \mathrm { s } }$