If `veca=hati+hatj+hatk, vecb=4hati+3hatj+4hatk` and `vecc=hati+alphahatj+betahatk`
are linearly dependent vectors and `|vecc|=sqrt(3)` then:

Options

A. (a) `alpha=1,beta=-1`
B. (b) `alpha=1,beta=+-1`
C. (c) `alpha=+-1,beta=+-1`
D. (d) `alpha=+-1,beta=1`

Explanation

To solve the problem, we will follow these steps:

### Step 1: Understand Linear Dependence
Given vectors \(\vec{a} = \hat{i} + \hat{j} + \hat{k}\), \(\vec{b} = 4\hat{i} + 3\hat{j} + 4\hat{k}\), and \(\vec{c} = \hat{i} + \alpha \hat{j} + \beta \hat{k}\), we know that these vectors are linearly dependent. This means that there exist scalars \(k_1\) and \(k_2\) such that:
\[ \vec{c} = k_1 \vec{a} + k_2 \vec{b} \]

### Step 2: Use the Box Product Condition
For three vectors to be linearly dependent, the scalar triple product (or box product) must be zero:
\[ \vec{a} \cdot (\vec{b} \times \vec{c}) = 0 \]
This can be represented as a determinant:
\[ \begin{vmatrix} 1 & 1 & 1 \\ 4 & 3 & 4 \\ 1 & \alpha & \beta \end{vmatrix} = 0 \]

### Step 3: Calculate the Determinant
Calculating the determinant:
\[ = 1 \cdot (3\beta - 4\alpha) - 1 \cdot (4\beta - 4) + 1 \cdot (4\alpha - 3) \]
This simplifies to:
\[ 3\beta - 4\alpha - 4\beta + 4 + 4\alpha - 3 = 0 \]
Combining like terms:
\[ (3\beta - 4\beta) + (-4\alpha + 4\alpha) + (4 - 3) = 0 \]
This gives:
\[ -\beta + 1 = 0 \implies \beta = 1 \]

### Step 4: Use the Magnitude Condition
We are given that the magnitude of \(\vec{c}\) is \(\sqrt{3}\):
\[ |\vec{c}| = \sqrt{1^2 + \alpha^2 + \beta^2} = \sqrt{3} \]
Squaring both sides:
\[ 1 + \alpha^2 + \beta^2 = 3 \]
Substituting \(\beta = 1\):
\[ 1 + \alpha^2 + 1^2 = 3 \implies 1 + \alpha^2 + 1 = 3 \]
This simplifies to:
\[ \alpha^2 + 2 = 3 \implies \alpha^2 = 1 \]
Thus, we have:
\[ \alpha = \pm 1 \]

### Conclusion
The values of \(\alpha\) and \(\beta\) are:
\[ \alpha = \pm 1, \quad \beta = 1 \]