Let $$\vec{a}$$, $$\vec{b}$$ and $$\vec{c}$$ be three unit vectors such that $$|\vec{a} - \vec{b}|^2 + |\vec{a} - \vec{c}|^2 = 8$$. Then $$|\vec{a} + 2\vec{b}|^2 + |\vec{a} + 2\vec{c}|^2$$ is equal to ___________.

We are given that $$\vec{a},\;\vec{b},\;\vec{c}$$ are unit vectors, that is $$|\vec{a}| = |\vec{b}| = |\vec{c}| = 1.$$

The first relation in the question is

$$|\vec{a}-\vec{b}|^{2} + |\vec{a}-\vec{c}|^{2} = 8.$$

We begin by expanding each squared modulus with the dot‐product formula

$$|\vec{u}-\vec{v}|^{2} = (\vec{u}-\vec{v})\!\cdot\!(\vec{u}-\vec{v}) = |\vec{u}|^{2} + |\vec{v}|^{2} - 2\,\vec{u}\!\cdot\!\vec{v}.$$

Applying this to $$\vec{u}=\vec{a},\;\vec{v}=\vec{b}$$ we get

$$|\vec{a}-\vec{b}|^{2} = |\vec{a}|^{2} + |\vec{b}|^{2} - 2\,\vec{a}\!\cdot\!\vec{b}.$$

Because the vectors are unit, $$|\vec{a}|^{2}=|\vec{b}|^{2}=1,$$ so

$$|\vec{a}-\vec{b}|^{2} = 1 + 1 - 2\,\vec{a}\!\cdot\!\vec{b} = 2\bigl(1-\vec{a}\!\cdot\!\vec{b}\bigr).$$

In the same manner, for $$\vec{u}=\vec{a},\;\vec{v}=\vec{c}$$ we obtain

$$|\vec{a}-\vec{c}|^{2} = 2\bigl(1-\vec{a}\!\cdot\!\vec{c}\bigr).$$

Now we add these two results exactly as the question does:

$$|\vec{a}-\vec{b}|^{2} + |\vec{a}-\vec{c}|^{2} = 2\bigl(1-\vec{a}\!\cdot\!\vec{b}\bigr) + 2\bigl(1-\vec{a}\!\cdot\!\vec{c}\bigr).$$

Combining like terms, this is

$$2\Bigl[\,2 - (\vec{a}\!\cdot\!\vec{b} + \vec{a}\!\cdot\!\vec{c})\Bigr].$$

The question tells us that this sum equals $$8,$$ so we equate:

$$2\Bigl[\,2 - (\vec{a}\!\cdot\!\vec{b} + \vec{a}\!\cdot\!\vec{c})\Bigr] = 8.$$

Dividing both sides by $$2$$ gives

$$2 - (\vec{a}\!\cdot\!\vec{b} + \vec{a}\!\cdot\!\vec{c}) = 4.$$

Transposing to isolate the dot‐product sum, we get

$$(\vec{a}\!\cdot\!\vec{b} + \vec{a}\!\cdot\!\vec{c}) = 2 - 4 = -2.$$

So we have established the key relation

$$\vec{a}\!\cdot\!\vec{b} + \vec{a}\!\cdot\!\vec{c} = -2.$$

Now we must evaluate $$|\vec{a}+2\vec{b}|^{2} + |\vec{a}+2\vec{c}|^{2}.$$

First expand $$|\vec{a}+2\vec{b}|^{2}$$ with the same dot‐product rule:

$$|\vec{a}+2\vec{b}|^{2} = (\vec{a}+2\vec{b})\!\cdot\!(\vec{a}+2\vec{b}) = |\vec{a}|^{2} + 4|\vec{b}|^{2} + 4\,\vec{a}\!\cdot\!\vec{b}.$$

Because $$|\vec{a}|^{2}=|\vec{b}|^{2}=1,$$ this simplifies to

$$|\vec{a}+2\vec{b}|^{2} = 1 + 4 + 4\,\vec{a}\!\cdot\!\vec{b} = 5 + 4\,\vec{a}\!\cdot\!\vec{b}.$$

Next, do the same for $$|\vec{a}+2\vec{c}|^{2}:$$

$$|\vec{a}+2\vec{c}|^{2} = (\vec{a}+2\vec{c})\!\cdot\!(\vec{a}+2\vec{c}) = |\vec{a}|^{2} + 4|\vec{c}|^{2} + 4\,\vec{a}\!\cdot\!\vec{c}.$$

Again inserting $$|\vec{a}|^{2}=|\vec{c}|^{2}=1,$$ we find

$$|\vec{a}+2\vec{c}|^{2} = 1 + 4 + 4\,\vec{a}\!\cdot\!\vec{c} = 5 + 4\,\vec{a}\!\cdot\!\vec{c}.$$

We now add these two expressions exactly as required:

$$|\vec{a}+2\vec{b}|^{2} + |\vec{a}+2\vec{c}|^{2} = \bigl(5 + 4\,\vec{a}\!\cdot\!\vec{b}\bigr) + \bigl(5 + 4\,\vec{a}\!\cdot\!\vec{c}\bigr).$$

Combining the constant terms and the dot‐product terms separately yields

$$10 + 4\bigl(\vec{a}\!\cdot\!\vec{b} + \vec{a}\!\cdot\!\vec{c}\bigr).$$

We have already found that $$\vec{a}\!\cdot\!\vec{b} + \vec{a}\!\cdot\!\vec{c} = -2,$$ so substituting this value gives

$$10 + 4(-2) = 10 - 8 = 2.$$

Thus, the numerical value requested by the question is

$$2.$$

So, the answer is $$2$$.