Let $$\overrightarrow{a}=\hat{i}+\hat{j}+\hat{k},\overrightarrow{b}=2\hat{i}+2\hat{j}+\hat{k}$$ and $$\overrightarrow{d}=\overrightarrow{a}\times \overrightarrow{b}$$. If$$\overrightarrow{c}$$ is a vector such that $$\overrightarrow{a}. \overrightarrow{c}=|\overrightarrow{c}|,|\overrightarrow{c}-2\overrightarrow{a}|^{2}=8$$ and the angle between $$\overrightarrow{d}$$ and $$\overrightarrow{c}$$ is $$\frac{\pi}{4}$$, then $$|10-3\overrightarrow{b}.\overrightarrow{c}|+|\overrightarrow{d}\times \overrightarrow{c}|^{2}$$ is equal to

Given vectors:

$$\overrightarrow{a} = \hat{i} + \hat{j} + \hat{k}$$

$$\overrightarrow{b} = 2\hat{i} + 2\hat{j} + \hat{k}$$

First, compute $$\overrightarrow{d} = \overrightarrow{a} \times \overrightarrow{b}$$.

The cross product is:

$$\overrightarrow{d} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 1 \\ 2 & 2 & 1 \end{vmatrix} = \hat{i}(1 \cdot 1 - 1 \cdot 2) - \hat{j}(1 \cdot 1 - 1 \cdot 2) + \hat{k}(1 \cdot 2 - 1 \cdot 2) = \hat{i}(-1) - \hat{j}(-1) + \hat{k}(0) = -\hat{i} + \hat{j}$$

So, $$\overrightarrow{d} = -\hat{i} + \hat{j}$$.

Let $$\overrightarrow{c} = x\hat{i} + y\hat{j} + z\hat{k}$$. The given conditions are:

1. $$\overrightarrow{a} \cdot \overrightarrow{c} = |\overrightarrow{c}|$$

$$\overrightarrow{a} \cdot \overrightarrow{c} = x + y + z$$

$$|\overrightarrow{c}| = \sqrt{x^2 + y^2 + z^2}$$

So,

$$x + y + z = \sqrt{x^2 + y^2 + z^2} \quad \text{(1)}$$

2. $$|\overrightarrow{c} - 2\overrightarrow{a}|^2 = 8$$

$$2\overrightarrow{a} = 2\hat{i} + 2\hat{j} + 2\hat{k}$$

$$\overrightarrow{c} - 2\overrightarrow{a} = (x - 2)\hat{i} + (y - 2)\hat{j} + (z - 2)\hat{k}$$

So,

$$(x - 2)^2 + (y - 2)^2 + (z - 2)^2 = 8 \quad \text{(2)}$$

3. Angle between $$\overrightarrow{d}$$ and $$\overrightarrow{c}$$ is $$\frac{\pi}{4}$$.

$$\overrightarrow{d} \cdot \overrightarrow{c} = (-1)(x) + (1)(y) + (0)(z) = -x + y$$

$$|\overrightarrow{d}| = \sqrt{(-1)^2 + 1^2} = \sqrt{2}$$

$$\cos \frac{\pi}{4} = \frac{\sqrt{2}}{2}$$

So,

$$\overrightarrow{d} \cdot \overrightarrow{c} = |\overrightarrow{d}| |\overrightarrow{c}| \cos \frac{\pi}{4} \implies -x + y = \sqrt{2} \cdot |\overrightarrow{c}| \cdot \frac{\sqrt{2}}{2} = |\overrightarrow{c}|$$

Thus,

$$-x + y = |\overrightarrow{c}| \quad \text{(3)}$$

From equations (1) and (3), let $$|\overrightarrow{c}| = k$$:

$$x + y + z = k \quad \text{(1)}$$

$$-x + y = k \quad \text{(3)}$$

Subtract equation (3) from equation (1):

$$(x + y + z) - (-x + y) = k - k \implies 2x + z = 0 \quad \text{(4)}$$

So, $$z = -2x$$.

From equation (3):

$$y = k + x \quad \text{(5)}$$

Substitute $$z = -2x$$ and $$y = k + x$$ into equation (1):

$$x + (k + x) + (-2x) = k \implies k = k$$ (always true).

Now, $$k = |\overrightarrow{c}| = \sqrt{x^2 + y^2 + z^2} = \sqrt{x^2 + (k + x)^2 + (-2x)^2}$$:

$$k^2 = x^2 + (k + x)^2 + (-2x)^2 = x^2 + k^2 + 2kx + x^2 + 4x^2 = k^2 + 6x^2 + 2kx$$

$$0 = 6x^2 + 2kx \implies 2x(3x + k) = 0$$

So, $$x = 0$$ or $$k = -3x$$.

**Case 1: $$x = 0$$**

From equation (4): $$z = -2(0) = 0$$.

From equation (5): $$y = k + 0 = k$$.

From equation (1): $$0 + k + 0 = k \implies k = k$$ (true).

Equation (2):

$$(0 - 2)^2 + (k - 2)^2 + (0 - 2)^2 = 4 + (k - 2)^2 + 4 = (k - 2)^2 + 8 = 8$$

$$(k - 2)^2 = 0 \implies k = 2$$

Then $$y = 2$$, so $$\overrightarrow{c} = 2\hat{j}$$.

**Case 2: $$k = -3x$$**

Since $$k \geq 0$$, $$-3x \geq 0 \implies x \leq 0$$.

From equation (5): $$y = -3x + x = -2x$$.

From equation (4): $$z = -2x$$.

Equation (2):

$$(x - 2)^2 + (-2x - 2)^2 + (-2x - 2)^2 = (x - 2)^2 + 2(-2(x + 1))^2 = (x - 2)^2 + 8(x + 1)^2 = 8$$

$$(x - 2)^2 + 8(x + 1)^2 = 8$$

$$x^2 - 4x + 4 + 8(x^2 + 2x + 1) = 8 \implies x^2 - 4x + 4 + 8x^2 + 16x + 8 = 8 \implies 9x^2 + 12x + 12 = 8$$

$$9x^2 + 12x + 4 = 0$$

Discriminant: $$12^2 - 4 \cdot 9 \cdot 4 = 144 - 144 = 0$$

$$x = \frac{-12}{18} = -\frac{2}{3}$$

Then $$k = -3 \left(-\frac{2}{3}\right) = 2$$, $$y = -2 \left(-\frac{2}{3}\right) = \frac{4}{3}$$, $$z = -2 \left(-\frac{2}{3}\right) = \frac{4}{3}$$, so $$\overrightarrow{c} = -\frac{2}{3}\hat{i} + \frac{4}{3}\hat{j} + \frac{4}{3}\hat{k}$$.

Now compute $$|10 - 3\overrightarrow{b} \cdot \overrightarrow{c}| + |\overrightarrow{d} \times \overrightarrow{c}|^2$$ for both cases.

**For $$\overrightarrow{c} = 2\hat{j}$$:**

$$\overrightarrow{b} \cdot \overrightarrow{c} = (2)(0) + (2)(2) + (1)(0) = 4$$

$$3\overrightarrow{b} \cdot \overrightarrow{c} = 12$$

$$10 - 12 = -2 \implies | -2 | = 2$$

$$\overrightarrow{d} \times \overrightarrow{c} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -1 & 1 & 0 \\ 0 & 2 & 0 \end{vmatrix} = \hat{i}(0) - \hat{j}(0) + \hat{k}(-2) = -2\hat{k}$$

$$|\overrightarrow{d} \times \overrightarrow{c}|^2 = (-2)^2 = 4$$

Sum: $$2 + 4 = 6$$.

**For $$\overrightarrow{c} = -\frac{2}{3}\hat{i} + \frac{4}{3}\hat{j} + \frac{4}{3}\hat{k}$$:**

$$\overrightarrow{b} \cdot \overrightarrow{c} = (2)\left(-\frac{2}{3}\right) + (2)\left(\frac{4}{3}\right) + (1)\left(\frac{4}{3}\right) = -\frac{4}{3} + \frac{8}{3} + \frac{4}{3} = \frac{8}{3}$$

$$3\overrightarrow{b} \cdot \overrightarrow{c} = 8$$

$$10 - 8 = 2 \implies |2| = 2$$

$$\overrightarrow{d} \times \overrightarrow{c} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -1 & 1 & 0 \\ -\frac{2}{3} & \frac{4}{3} & \frac{4}{3} \end{vmatrix} = \hat{i}\left(1 \cdot \frac{4}{3} - 0 \cdot \frac{4}{3}\right) - \hat{j}\left(-1 \cdot \frac{4}{3} - 0 \cdot \left(-\frac{2}{3}\right)\right) + \hat{k}\left(-1 \cdot \frac{4}{3} - 1 \cdot \left(-\frac{2}{3}\right)\right) = \frac{4}{3}\hat{i} + \frac{4}{3}\hat{j} - \frac{2}{3}\hat{k}$$

$$|\overrightarrow{d} \times \overrightarrow{c}|^2 = \left(\frac{4}{3}\right)^2 + \left(\frac{4}{3}\right)^2 + \left(-\frac{2}{3}\right)^2 = \frac{16}{9} + \frac{16}{9} + \frac{4}{9} = \frac{36}{9} = 4$$

Sum: $$2 + 4 = 6$$.

In both cases, the expression evaluates to 6.

Thus, $$|10 - 3\overrightarrow{b} \cdot \overrightarrow{c}| + |\overrightarrow{d} \times \overrightarrow{c}|^2 = 6$$.

**Final Answer:** $$6$$