`vecp=3hati+2hatj+hatk ,vecq=2hati+hatj+hatk, vecr` is `bot` to both `vecp+vecq and (vecp-vecq), absvecr=sqrt3. vecr=ahati+bhatj+chatk.` then find `absveca + absvecb + absvecc`.

Answer

To solve the problem step by step, we start with the given vectors and the conditions provided.

### Step 1: Define the vectors
We are given:
\[ \vec{p} = 3\hat{i} + 2\hat{j} + \hat{k} \]
\[ \vec{q} = 2\hat{i} + \hat{j} + \hat{k} \]

### Step 2: Calculate \(\vec{p} + \vec{q}\) and \(\vec{p} - \vec{q}\)
First, we find:
\[ \vec{p} + \vec{q} = (3 + 2)\hat{i} + (2 + 1)\hat{j} + (1 + 1)\hat{k} = 5\hat{i} + 3\hat{j} + 2\hat{k} \]
\[ \vec{p} - \vec{q} = (3 - 2)\hat{i} + (2 - 1)\hat{j} + (1 - 1)\hat{k} = 1\hat{i} + 1\hat{j} + 0\hat{k} \]

### Step 3: Find the cross product \((\vec{p} + \vec{q}) \times (\vec{p} - \vec{q})\)
We need to compute the cross product:
\[ \vec{a} = \vec{p} + \vec{q} = 5\hat{i} + 3\hat{j} + 2\hat{k} \]
\[ \vec{b} = \vec{p} - \vec{q} = 1\hat{i} + 1\hat{j} + 0\hat{k} \]

The cross product \(\vec{a} \times \vec{b}\) is given by the determinant:
\[ \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 5 & 3 & 2 \\ 1 & 1 & 0 \end{vmatrix} \]

Calculating this determinant:
\[ \hat{i} \begin{vmatrix} 3 & 2 \\ 1 & 0 \end{vmatrix} - \hat{j} \begin{vmatrix} 5 & 2 \\ 1 & 0 \end{vmatrix} + \hat{k} \begin{vmatrix} 5 & 3 \\ 1 & 1 \end{vmatrix} \]

Calculating each of the 2x2 determinants:
1. \(\begin{vmatrix} 3 & 2 \\ 1 & 0 \end{vmatrix} = (3 \cdot 0 - 2 \cdot 1) = -2\)
2. \(\begin{vmatrix} 5 & 2 \\ 1 & 0 \end{vmatrix} = (5 \cdot 0 - 2 \cdot 1) = -2\)
3. \(\begin{vmatrix} 5 & 3 \\ 1 & 1 \end{vmatrix} = (5 \cdot 1 - 3 \cdot 1) = 2\)

Thus, we have:
\[ \vec{a} \times \vec{b} = -2\hat{i} + 2\hat{j} + 2\hat{k} \]

### Step 4: Express \(\vec{r}\) in terms of \(\lambda\)
Since \(\vec{r}\) is perpendicular to both \(\vec{a}\) and \(\vec{b}\), we can express \(\vec{r}\) as:
\[ \vec{r} = \lambda (-2\hat{i} + 2\hat{j} + 2\hat{k}) \]
This means:
\[ \vec{r} = -2\lambda \hat{i} + 2\lambda \hat{j} + 2\lambda \hat{k} \]

### Step 5: Find the magnitude of \(\vec{r}\)
The magnitude of \(\vec{r}\) is given as:
\[ |\vec{r}| = \sqrt{(-2\lambda)^2 + (2\lambda)^2 + (2\lambda)^2} = \sqrt{4\lambda^2 + 4\lambda^2 + 4\lambda^2} = \sqrt{12\lambda^2} = 2\sqrt{3}|\lambda| \]
We know that \(|\vec{r}| = \sqrt{3}\), hence:
\[ 2\sqrt{3}|\lambda| = \sqrt{3} \]
This gives:
\[ |\lambda| = \frac{1}{2} \]

### Step 6: Substitute \(\lambda\) back into \(\vec{r}\)
Now substituting \(\lambda\) back into \(\vec{r}\):
\[ \vec{r} = -2\left(\frac{1}{2}\right)\hat{i} + 2\left(\frac{1}{2}\right)\hat{j} + 2\left(\frac{1}{2}\right)\hat{k} = -\hat{i} + \hat{j} + \hat{k} \]

### Step 7: Identify components \(a\), \(b\), and \(c\)
From \(\vec{r} = a\hat{i} + b\hat{j} + c\hat{k}\), we have:
\[ a = -1, \quad b = 1, \quad c = 1 \]

### Step 8: Calculate the absolute values and their sum
Now we calculate:
\[ |\vec{a}| + |\vec{b}| + |\vec{c}| = | -1 | + | 1 | + | 1 | = 1 + 1 + 1 = 3 \]

### Final Answer
Thus, the final answer is:
\[ \boxed{3} \]