Answer
${F}^{>}_{3}$=-$7i$-$12j$ $N$
Work Step by Step
Let $F^{>}_{net}$ be the net force which produces acceleration so
$F^{>}_{net}$=$F^{>}_{1}$+$F^{>}_{2}$+$F^{>}_{3}$
As we know that
$F^{>}_{net}$=$ma^{>}$
Hence
$ma^{>}$=$F^{>}_{1}$+$F^{>}_{2}$+$F^{>}_{3}$
$ma^{>}$-$F^{>}_{1}$-$F^{>}_{2}$=$F^{>}_{3}$
or
$F^{>}_{3}$=$ma^{>}$-$F^{>}_{1}$-$F^{>}_{2}$
Putting the values, you will get
$F^{>}_{3}$=$2(-8i+6j)-(3i+16j)-(-12i+8j)$
$F^{>}_{3}$=$-16i+12j-3i-16j+12i-8j$
$F^{>}_{3}$=$-7i-12j$ $N$
