First, applying your equation to your case, $n_1-(n_2+n_3) + 0 - 0 = 0$ or $n_1 = n_2+n_3$
$$n_{1,a}=n_{2,a}+n_{3,a} \space (1) \\
n_{1,b}=n_{2,b}+n_{3,b} \space (2)\\
n_{1,c}=n_{2,c}+n_{3,c} \space (3)
$$
and since there's nothing that separates/filters the components in different proportions, the output will contain exactly the same fraction as input.
$$ f n_{2,a} + (1-f) n_{3,a} = n_{1,a} \space (4)\\
f n_{2,b} + (1-f) n_{3,b} = n_{1,b} \space (5)\\
f n_{2,c} + (1-f) n_{3,c} = n_{1,c} \space (6)\\
$$
where $f$ is the ratio at which the streams are split into output - the same for all components.
Looking at the table, we have $n_{1,A}$ and $n_{2,a}$. From (1) we can quickly find $n_{3,a}$ : 4-1=3.
With these 3 we can find $f$ from (4): 0.25
Remainder of $n$ is trivial as we have column $1$ - multiply the $1$ column value by 0.25 for column $2$ and 0.75 for column $3$, e.g. $n_{2,b} = 10\cdot0.25 = 2.5$
For totals in kg/h, simply take the units given to find the factor by which to multiply:
$$total_{c} = \sum_i^{a,b,c}{MW_i \cdot n_{c,i}} $$
(remembering the 1000 factor of $kmol$ vs $mol$ - multiply $n$ by 1000 first): For column $1$
(4*10 + 10*20 + 6*30)*1,000 = 40+200+180 = 420,000 kg/h.