Let $$\vec{p} = 2\hat{i} + 3\hat{j} + \hat{k}$$ and $$\vec{q} = \hat{i} + 2\hat{j} + \hat{k}$$ be two vectors. If a vector $$\vec{r} = \alpha\hat{i} + \beta\hat{j} + \gamma\hat{k}$$ is perpendicular to each of the vectors $$(\vec{p} + \vec{q})$$ and $$(\vec{p} - \vec{q})$$, and $$|\vec{r}| = \sqrt{3}$$, then $$|\alpha| + |\beta| + |\gamma|$$ is equal to ___.

We have the vectors $$\vec{p}=2\hat{i}+3\hat{j}+\hat{k}$$ and $$\vec{q}=\hat{i}+2\hat{j}+\hat{k}$$. First we form the two combinations that are mentioned in the problem.

Adding the two given vectors, we obtain $$\vec{p}+\vec{q}=(2\hat{i}+\hat{i})+(3\hat{j}+2\hat{j})+(\hat{k}+\hat{k}) =3\hat{i}+5\hat{j}+2\hat{k}.$$

Subtracting them, we get $$\vec{p}-\vec{q}=(2\hat{i}-\hat{i})+(3\hat{j}-2\hat{j})+(\hat{k}-\hat{k}) =\hat{i}+\hat{j}+0\hat{k}.$$

The unknown vector is given as $$\vec{r}=\alpha\hat{i}+\beta\hat{j}+\gamma\hat{k}.$$ It is stated that $$\vec{r}$$ is perpendicular to each of the two vectors found above. The fundamental fact we use is: Two vectors are perpendicular if and only if their dot product is zero.

Applying this to $$\vec{r}$$ and $$\vec{p}+\vec{q}$$, we write $$\vec{r}\cdot(\vec{p}+\vec{q})=0.$$ Using the component-wise formula for the dot product, $$\vec{r}\cdot(\vec{p}+\vec{q}) =(\alpha\hat{i}+\beta\hat{j}+\gamma\hat{k})\cdot(3\hat{i}+5\hat{j}+2\hat{k}) =3\alpha+5\beta+2\gamma=0.$$ So we have our first linear equation: $$3\alpha+5\beta+2\gamma=0.\qquad(1)$$

Next we impose perpendicularity between $$\vec{r}$$ and $$\vec{p}-\vec{q}$$. Thus $$\vec{r}\cdot(\vec{p}-\vec{q})=0,$$ which gives $$(\alpha\hat{i}+\beta\hat{j}+\gamma\hat{k})\cdot(\hat{i}+\hat{j}+0\hat{k}) =\alpha+\beta+0\gamma=0.$$ Hence the second linear equation is $$\alpha+\beta=0.\qquad(2)$$

From equation (2) we can immediately express $$\beta$$ in terms of $$\alpha$$: $$\beta=-\alpha.$$

Substituting $$\beta=-\alpha$$ into equation (1), we obtain $$3\alpha+5(-\alpha)+2\gamma=0 \;\;\Longrightarrow\;\; 3\alpha-5\alpha+2\gamma=0 \;\;\Longrightarrow\;\; -2\alpha+2\gamma=0 \;\;\Longrightarrow\;\; \gamma=\alpha.$$

At this stage we have expressed every component of $$\vec{r}$$ in terms of the single parameter $$\alpha$$: $$\vec{r}=\alpha\hat{i}+(-\alpha)\hat{j}+\alpha\hat{k} =\alpha\big(\hat{i}-\hat{j}+\hat{k}\big).$$

The length of $$\vec{r}$$ is prescribed to be $$\sqrt{3}$$. The magnitude of a scalar multiple obeys $$|\lambda\vec{u}|=|\lambda|\,|\vec{u}|$$. Therefore $$|\vec{r}|=|\alpha|\,\big|\hat{i}-\hat{j}+\hat{k}\big|=\sqrt{3}.$$

We now compute the magnitude of the vector $$\hat{i}-\hat{j}+\hat{k}$$: $$\big|\hat{i}-\hat{j}+\hat{k}\big| =\sqrt{(1)^2+(-1)^2+(1)^2} =\sqrt{1+1+1} =\sqrt{3}.$$

Putting this value into the length condition, we get $$|\alpha|\,\sqrt{3}=\sqrt{3} \;\;\Longrightarrow\;\; |\alpha|=1.$$

Because $$\beta=-\alpha$$ and $$\gamma=\alpha$$, we have $$|\beta|=|\alpha|=1,\qquad|\gamma|=|\alpha|=1.$$

Finally, the required sum is $$|\alpha|+|\beta|+|\gamma|=1+1+1=3.$$

So, the answer is $$3$$.