1. Given vectors

$\vec p = 2\hat i + 3\hat j + \hat k = (2,3,1), \qquad \vec q = \hat i + 2\hat j + \hat k = (1,2,1)$

We are told that

$\vec r = \alpha \hat i + \beta \hat j + \gamma \hat k = (\alpha,\beta,\gamma)$

is perpendicular to both $\vec p+\vec q$ and $\vec p-\vec q$.

2. Compute $\vec p+\vec q$ and $\vec p-\vec q$

$\vec p+\vec q = (2+1,\,3+2,\,1+1) = (3,5,2)$

$\vec p-\vec q = (2-1,\,3-2,\,1-1) = (1,1,0)$

Since $\vec r$ is perpendicular to both, we must have

$\vec r\cdot (3,5,2)=0$ $\vec r\cdot (1,1,0)=0$

So,

$3\alpha+5\beta+2\gamma=0 \quad \cdots (1)$ $\alpha+\beta=0 \quad \cdots (2)$

3. Solve for the direction of $\vec r$

From (2),

$\alpha=-\beta$

Substitute into (1):

$3(-\beta)+5\beta+2\gamma=0$ $2\beta+2\gamma=0$ $\beta+\gamma=0$ $\gamma=-\beta$

Thus,

$\alpha=-\beta, \qquad \gamma=-\beta$

Hence,

$\vec r = (-\beta,\beta,-\beta)=\beta(-1,1,-1)$

4. Use the magnitude condition

Given

$|\vec r|=\sqrt 3$

Now,

$|\vec r|=|\beta|\sqrt{(-1)^2+1^2+(-1)^2}=|\beta|\sqrt3$

So,

$|\beta|\sqrt3=\sqrt3$ $|\beta|=1$

Therefore,

$|\alpha|=1, \quad |\beta|=1, \quad |\gamma|=1$

5. Required value

$|\alpha|+|\beta|+|\gamma|=1+1+1=3$

6. Comparison with stored answer

Derived answer = $3$.

Stored correct answer = $3$.

They agree.