We are given a triangle $ABC$ with $AB = 6$, $AC = 3$, and $\angle BAC = 120^\circ$. $AD$ is an angle bisector of $\angle BAC$. We are asked to find the lengths of $AD$, $AE$, and the inradii $r_1$ and $r_2$ of triangles $ADC$ and $ADE$, respectively, as well as the angle $\angle DAE$. We are given that $BE = 3\sqrt{7}$ and $CD = \sqrt{7}$.

1. Problem Description

We are given a triangle $ABC$ with $AB = 6$, $AC = 3$, and $\angle BAC = 120^\circ$. $AD$ is an angle bisector of $\angle BAC$. We are asked to find the lengths of $AD$, $AE$, and the inradii $r_1$ and $r_2$ of triangles $ADC$ and $ADE$, respectively, as well as the angle $\angle DAE$. We are given that $BE = 3\sqrt{7}$ and $CD = \sqrt{7}$.

2. Solution Steps

First, we find $AD$ using the area relation $\triangle ABD + \triangle ACD = \triangle ABC$.

The area of $\triangle ABC$ is

$\frac{1}{2} \times AB \times AC \times \sin(120^\circ) = \frac{1}{2} \times 6 \times 3 \times \frac{\sqrt{3}}{2} = \frac{9\sqrt{3}}{2}$.

The area of $\triangle ABD$ is $\frac{1}{2} \times AB \times AD \times \sin(60^\circ) = \frac{1}{2} \times 6 \times AD \times \frac{\sqrt{3}}{2} = \frac{3\sqrt{3}}{2} AD$.

The area of $\triangle ACD$ is $\frac{1}{2} \times AC \times AD \times \sin(60^\circ) = \frac{1}{2} \times 3 \times AD \times \frac{\sqrt{3}}{2} = \frac{3\sqrt{3}}{4} AD$.

Then $\frac{3\sqrt{3}}{2} AD + \frac{3\sqrt{3}}{4} AD = \frac{9\sqrt{3}}{2}$.

$\frac{6\sqrt{3} + 3\sqrt{3}}{4} AD = \frac{9\sqrt{3}}{2}$.

$\frac{9\sqrt{3}}{4} AD = \frac{9\sqrt{3}}{2}$.

$AD = 2$.

So, $I=2$.

Next, we find $\angle DAE$. Since $AD$ is an angle bisector, $\angle BAD = \angle CAD = 60^\circ$. We have $CD = \sqrt{7}$, so $CE = BE - BC = 3\sqrt{7} - (2\sqrt{7} + \sqrt{7}) = 0$, which seems incorrect.

Since $BC = BD+DC=2\sqrt{7}+\sqrt{7}=3\sqrt{7}$ and $BE=3\sqrt{7}$, this means $E=C$.

Since $E=C$, $AE = AC = 3$ and $\angle DAE=0^\circ$.

Therefore, $\angle DAE = 0^\circ$ and $AE=3$. Thus $JK = 0$, $L = 3$, $M=1$.

To find $r_1$, we use the formula $r_1 = \frac{2 \times Area}{Perimeter}$ for $\triangle ADC$.

The area of $\triangle ADC$ is $\frac{3\sqrt{3}}{4}AD = \frac{3\sqrt{3}}{4} \times 2 = \frac{3\sqrt{3}}{2}$.

The perimeter of $\triangle ADC$ is $3 + \sqrt{7} + 2 = 5 + \sqrt{7}$.

$r_1 = \frac{2 \times \frac{3\sqrt{3}}{2}}{3 + \sqrt{7} + 2} = \frac{3\sqrt{3}}{5 + \sqrt{7}} = \frac{3\sqrt{3}(5 - \sqrt{7})}{(5 + \sqrt{7})(5 - \sqrt{7})} = \frac{15\sqrt{3} - 3\sqrt{21}}{25 - 7} = \frac{15\sqrt{3} - 3\sqrt{21}}{18} = \frac{5\sqrt{3} - \sqrt{21}}{6}$.

Thus $N=5$, $O=3$, $PQ=21$, $R=6$.

Since $E=C$, then $ADE$ does not exist, which implies $r_2$ doesn't exist.

If $E \ne C$, since $\angle DAE = 0^\circ$, $AD$ and $AE$ are on the same line, i.e., $AD$ is an extension of $AE$. But $AD$ bisects $\angle BAC$. Since $AC=3$ and $AE = 3$, then $C = E$.

Since the problem is not well-defined ($E=C$), we cannot proceed further.

Final Answer:

1. Problem Description

We are given a triangle $ABC$ with $AB = 6$, $AC = 3$, and $\angle BAC = 120^\circ$. $AD$ is an angle bisector of $\angle BAC$. $BE = 3\sqrt{7}$ and $CD = \sqrt{7}$.

The task is to find the lengths of $AD$, $AE$, and the inradii $r_1$ and $r_2$ of triangles $ADC$ and $ADE$, respectively, and the angle $\angle DAE$.

2. Solution Steps

$AD = 2$

Since $E=C$, $AE = 3$ and $\angle DAE = 0^\circ$

$r_1 = \frac{5\sqrt{3} - \sqrt{21}}{6}$

Since $E=C$, triangle ADE does not exist. Thus, $r_2$ cannot be calculated.

3. Final Answer

$I=2$

$JK=0$

$L=3$

$M=1$

$N=5$

$O=3$

$PQ=21$

$R=6$

The values for $S$, $T$, $U$, $V$, and $W$ cannot be determined as $E=C$.