m is divisible by 5 and less than 120
$$One\ of\ the\ roots\ of\ \ x^2-2px+m=0\ is\ 7$$
m is the product of roots
$$Let\ the\ \ other\ roots\ of\ x^2-2px+m=0\ be\ k.$$
$$Let\ the\ \ ot7\cdot k=m$$
$$m\ can\ be\ any\ value\ \ge0\ or\ \le115\ \sin ce\ it\ is\ divisible\ by\ 5\ and\ less\ than\ 120\ \ while\ a\ $$ $$So,\ in\ x^2-2px+m=0;\ x=7\ as\ one\ of\ it\ 's\ root\ \ $$
$$7^2-2px+m=0$$
$$49-14p+m=0$$
$$14p=49+m\ $$
p is expected to be the prime and the prime that makes m divisible by 5 and less than 120 eqivalent to 11.
So p =11 and m= 105

One of the roots of $$x^2-2p\left(12\right)+n\ =0\ \ is\ 12$$
So x = 12

$$12^2-2p\left(12\right)+n\ =0$$
$$144-24p+n=0\ where\ \ \ p\ =\ 11$$
$$144-24p+n=0$$
$$n=120$$

$$Therefore,\ p+n-m$$ $$11+120-105$$ $$11+120-105$$ $$=26$$ $$Ans\ is\ option\ D$$