The appropriate identity to factorise $49y^2+84yz+36z^2$ will be:

The given expression $49{\mathrm{y}}^2+84\mathrm{yz}+36{\mathrm{z}}^2$ has three terms, in which the first and the last terms are perfect squares. Also, all the terms are positive. 

Let us compare it to the identity ${(\mathrm{a}+\mathrm{b})}^2={\mathrm{a}}^2+2\mathrm{ab}+{\mathrm{b}}^2$. 

$49{\mathrm{y}}^2+84\mathrm{yz}+36{\mathrm{z}}^2$ can be written as:

${\left(7\mathrm{y}\right)}^2+(2\times 7\mathrm{y}\times 6\mathrm{z})+{\left(6\mathrm{z}\right)}^2$

$={(7\mathrm{y}+6\mathrm{z})}^2$

Hence, we can factorise it using the identity: ${(\mathrm{a}+\mathrm{b})}^2={\mathrm{a}}^2+2\mathrm{ab}+{\mathrm{b}}^2$.