$\because(1+x)^{2}=1+2 x+x^{2},$

$\left(1+x^{2}\right)^{3}=1+3 x^{2}+3 x^{4}+x^{6},$

and $\left(1+x^{3}\right)^{4}=1+4 x^{3}+6 x^{6}+4 x^{9}+x^{12}$

So, the possible combinations for $x^{10}$ are:

$x \cdot x^{9}, x \cdot x^{6} \cdot x^{3}, x^{2} \cdot x^{2} \cdot x^{6}, x^{4} \cdot x^{6}$

Corresponding coefficientsare $2 \times 4,2 \times 1 \times 4,1 \times 3 \times 6,3 \times 6$ 

or $8,8,18,18$

$\therefore$ Sum of the coefficient is

$8+8+18+18=52$

Therefore, the coefficient of $x^{10}$ in the expansion of $(1+x)^{2}\left(1+x^{2}\right)^{3}\left(1+x^{3}\right)^{4}$ is $52$