A degree 6 polynomial can have 6 real zeroes; 2 real zeroes and 4 complex zeroes; 4 complex zeroes and 2 real zeroes; 6 complex zeroes.
The given information shows one real zero only, so there must be another to make a pair.
Of the remaining 4 complex zeroes there have to be 2 conjugate pairs. I guess that 13i and -7a3i are conjugates, so -7a3i=-13i, 7a3=13, a3=13/7, a=1.2291 approx.
(x-13i)(x+13i)=x2+169.
-10+i√11 and -10-i√11 is a conjugate pair, so (x+10-i√11)(x+10+i√11)=x2+20x+221.
(x2+169)(x2+20x+221)=x4+20x3+390x2+3380x+37349.
x+12 is another factor, so:
x5+20x4+390x3+3380x2+37349x+12x4+240x3+4680x2+40560x+448188=
x5+32x4+630x3+8060x2+77909x+448188.
We need another factor, x-p, polynomial is:
x6+32x5+630x4+8060x3+77909x2+448188x-p(x5+32x4+630x3+8060x2+77909x+448188)=
x6+(32-p)x5+(630-32p)x4+(8060-630p)x3+(77909-8060p)x2+(448188-77909p)x+448188p, where p is a real number. There doesn't appear to be any way to find p, the missing zero, unless there is an error in the question.