Explanation

(b)
We have,
$(xy + yz + zx)^{6} = \sum_{r + s + t = 6}^{}{\frac{6!}{r!s!t!}(xy)^{r}(yz)^{s}(zx)^{t}}$
$= \sum_{r + s + t = 6}^{}{\frac{6!}{r!s!t!}x^{r + t}y^{r + s}z^{s + t}}$
If the general term in the above expansion contains $x^{3}y^{4}z^{5},$ then
$r + t = 3,\ r + s = 4$ and $s + t = 5$
Also, $r + s + t = 6$
On solving these equations, we get
$\ \ r = 1,\ s = 3,\ t = 2$
$\therefore Coefficient\ of\ x^{3}y^{4}z^{5} = \frac{6!}{1!3!2!} = 60$