Let $$\overrightarrow{r}=2\hat{i}-\hat{j}+3\hat{k}, \overrightarrow{c}=3\hat{i}-5\hat{j}+\hat{k}$$ and $$\overrightarrow{c}$$ be a vector such that $$\overrightarrow{c} \times \overrightarrow{c} = \overrightarrow{c} \times \overrightarrow{b}$$ and $$(\overrightarrow{a}+\overrightarrow{c}).(\overrightarrow{b}.\overrightarrow{c})=168$$. Then the maximum value of $$|\overrightarrow{c}|^{2}$$ is :

Given vectors are $$\overrightarrow{a} = 2\hat{i} - \hat{j} + 3\hat{k}$$ and $$\overrightarrow{b} = 3\hat{i} - 5\hat{j} + \hat{k}$$. The conditions are:

1. $$\overrightarrow{a} \times \overrightarrow{c} = \overrightarrow{c} \times \overrightarrow{b}$$
2. $$(\overrightarrow{a} + \overrightarrow{c}) \cdot (\overrightarrow{b} \cdot \overrightarrow{c}) = 168$$, but this appears to have a typo. Given the context and solution, it is interpreted as $$(\overrightarrow{a} + \overrightarrow{c}) \cdot (\overrightarrow{b} + \overrightarrow{c}) = 168$$.

From the first condition:

$$\overrightarrow{a} \times \overrightarrow{c} = \overrightarrow{c} \times \overrightarrow{b}$$
Since $$\overrightarrow{c} \times \overrightarrow{b} = - \overrightarrow{b} \times \overrightarrow{c}$$, we have:
$$\overrightarrow{a} \times \overrightarrow{c} = - \overrightarrow{b} \times \overrightarrow{c}$$
$$\overrightarrow{a} \times \overrightarrow{c} + \overrightarrow{b} \times \overrightarrow{c} = \overrightarrow{0}$$
$$(\overrightarrow{a} + \overrightarrow{b}) \times \overrightarrow{c} = \overrightarrow{0}$$
Thus, $$\overrightarrow{c}$$ is parallel to $$\overrightarrow{a} + \overrightarrow{b}$$.

Compute $$\overrightarrow{a} + \overrightarrow{b}$$:
$$\overrightarrow{a} + \overrightarrow{b} = (2 + 3)\hat{i} + (-1 - 5)\hat{j} + (3 + 1)\hat{k} = 5\hat{i} - 6\hat{j} + 4\hat{k}$$
Let $$\overrightarrow{d} = 5\hat{i} - 6\hat{j} + 4\hat{k}$$, so $$\overrightarrow{c} = k \overrightarrow{d} = k(5\hat{i} - 6\hat{j} + 4\hat{k})$$ for some scalar $$k$$.

Now, the second condition is $$(\overrightarrow{a} + \overrightarrow{c}) \cdot (\overrightarrow{b} + \overrightarrow{c}) = 168$$.
Substitute $$\overrightarrow{c} = k(5\hat{i} - 6\hat{j} + 4\hat{k})$$:
$$\overrightarrow{a} + \overrightarrow{c} = (2 + 5k)\hat{i} + (-1 - 6k)\hat{j} + (3 + 4k)\hat{k}$$
$$\overrightarrow{b} + \overrightarrow{c} = (3 + 5k)\hat{i} + (-5 - 6k)\hat{j} + (1 + 4k)\hat{k}$$

Dot product:
$$(\overrightarrow{a} + \overrightarrow{c}) \cdot (\overrightarrow{b} + \overrightarrow{c}) = (2 + 5k)(3 + 5k) + (-1 - 6k)(-5 - 6k) + (3 + 4k)(1 + 4k)$$
Expand each term:
$$(2 + 5k)(3 + 5k) = 2 \cdot 3 + 2 \cdot 5k + 5k \cdot 3 + 5k \cdot 5k = 6 + 10k + 15k + 25k^2 = 6 + 25k + 25k^2$$
$$(-1 - 6k)(-5 - 6k) = (-1)(-5) + (-1)(-6k) + (-6k)(-5) + (-6k)(-6k) = 5 + 6k + 30k + 36k^2 = 5 + 36k + 36k^2$$
$$(3 + 4k)(1 + 4k) = 3 \cdot 1 + 3 \cdot 4k + 4k \cdot 1 + 4k \cdot 4k = 3 + 12k + 4k + 16k^2 = 3 + 16k + 16k^2$$

Sum the expressions:
$$(6 + 5 + 3) + (25k + 36k + 16k) + (25k^2 + 36k^2 + 16k^2) = 14 + 77k + 77k^2$$

Set equal to 168:
$$77k^2 + 77k + 14 = 168$$
$$77k^2 + 77k - 154 = 0$$
Divide by 77:
$$k^2 + k - 2 = 0$$
Solve the quadratic equation:
Discriminant $$D = 1^2 - 4 \cdot 1 \cdot (-2) = 1 + 8 = 9$$
$$k = \frac{-1 \pm \sqrt{9}}{2} = \frac{-1 \pm 3}{2}$$
So, $$k = \frac{-1 + 3}{2} = 1$$ or $$k = \frac{-1 - 3}{2} = -2$$

Now, $$|\overrightarrow{c}|^2 = k^2 |\overrightarrow{d}|^2$$
$$|\overrightarrow{d}|^2 = 5^2 + (-6)^2 + 4^2 = 25 + 36 + 16 = 77$$
For $$k = 1$$, $$|\overrightarrow{c}|^2 = (1)^2 \cdot 77 = 77$$
For $$k = -2$$, $$|\overrightarrow{c}|^2 = (-2)^2 \cdot 77 = 4 \cdot 77 = 308$$
The maximum value is 308.

Verify the second condition for both values:
For $$k = 1$$, $$\overrightarrow{c} = 5\hat{i} - 6\hat{j} + 4\hat{k}$$
$$\overrightarrow{a} + \overrightarrow{c} = (2 + 5)\hat{i} + (-1 - 6)\hat{j} + (3 + 4)\hat{k} = 7\hat{i} - 7\hat{j} + 7\hat{k}$$
$$\overrightarrow{b} + \overrightarrow{c} = (3 + 5)\hat{i} + (-5 - 6)\hat{j} + (1 + 4)\hat{k} = 8\hat{i} - 11\hat{j} + 5\hat{k}$$
Dot product: $$7 \cdot 8 + (-7) \cdot (-11) + 7 \cdot 5 = 56 + 77 + 35 = 168$$
For $$k = -2$$, $$\overrightarrow{c} = -10\hat{i} + 12\hat{j} - 8\hat{k}$$
$$\overrightarrow{a} + \overrightarrow{c} = (2 - 10)\hat{i} + (-1 + 12)\hat{j} + (3 - 8)\hat{k} = -8\hat{i} + 11\hat{j} - 5\hat{k}$$
$$\overrightarrow{b} + \overrightarrow{c} = (3 - 10)\hat{i} + (-5 + 12)\hat{j} + (1 - 8)\hat{k} = -7\hat{i} + 7\hat{j} - 7\hat{k}$$
Dot product: $$(-8) \cdot (-7) + 11 \cdot 7 + (-5) \cdot (-7) = 56 + 77 + 35 = 168$$

Both satisfy the second condition, and the first condition is satisfied since $$\overrightarrow{c}$$ is parallel to $$\overrightarrow{a} + \overrightarrow{b}$$ for both $$k$$. The maximum value of $$|\overrightarrow{c}|^2$$ is 308.

Final Answer: 308