Incenter Circle

We can clearly notice from diagram that-

$\angle IC_1B = 180 - \angle B -\frac{\angle C}{2}$

$\angle IB_1A = 180 - \angle A -\frac{\angle B}{2}$

Since $AB_1IC_1$ is $cyclic$ , $m\angle IC_1B = m\angle IB_1A$

$\Rightarrow 180 - \angle B -\frac{\angle C}{2} = 180 - \angle A -\frac{\angle B}{2}$

$\Rightarrow A = \frac{B + C}{2}$

Also,

$A + B + C = 180 \Rightarrow \frac{B + C}{2} + B + C = 180$

$\Rightarrow B + C = 120 \Rightarrow A = 60$

Using Extension of Pythagoras theorem for A = 60 ,

$(BC)^2 = (11)^2 + (7)^2 - (11)(7) = 121 + 49 - 77 = 93$

$\Rightarrow \huge \boxed{n = 93}$

Imgur

In triangle CBI : $\alpha + \beta + \gamma = 180 = \gamma + \theta$

$\rightarrow \alpha + \beta = \theta$

In triangle ABC: $2\alpha + 2\beta + \theta = 3\alpha + 3\beta = 180$

$\rightarrow \alpha + \beta = \theta = 60$

$\cos \theta = \cos 60 = \frac{1}{2} = \frac{11^2 + 7^2 - n}{2 \times 11 \times 7}$

$\rightarrow n = 93$