Statement-I If `a=3hat(i)-3hat(j)+hat(k), b=-hat(i)+2hat(j)+hat(k) and c=hat(i)+hat(j)+hat(k) and d=2hat(i)-hat(j)`, then there exist real numbers `alpha, beta, gamma` such that `a=alphab+betac+gammad`
Statement-II a, b, c, d are four vectors in a 3-dimensional space. If b, c, d are non-coplanar, then there exist real numbers `alpha, beta, gamma` such that `a=alphab+betac+gammad.`

Options

A. Both Statement-I and Statement-II are correct and Statement-II is the correct explanation of Statement-I
B. Both Statement-I and Statement-II are correct but Statement-II is not the correct explanation of Statement-I
C. Statement-I is correct but Statement-II is incorrect
D. Statement-II is correct but Statement-I is incorrect

Answer

To solve the problem, we need to analyze the two statements provided and determine their validity based on the vectors given.

### Step 1: Identify the vectors
We have the following vectors:
- \( \mathbf{a} = 3\hat{i} - 3\hat{j} + \hat{k} \)
- \( \mathbf{b} = -\hat{i} + 2\hat{j} + \hat{k} \)
- \( \mathbf{c} = \hat{i} + \hat{j} + \hat{k} \)
- \( \mathbf{d} = 2\hat{i} - \hat{j} \)

### Step 2: Check if vectors \( \mathbf{b}, \mathbf{c}, \mathbf{d} \) are coplanar
To check if the vectors \( \mathbf{b}, \mathbf{c}, \mathbf{d} \) are coplanar, we can calculate the determinant of the matrix formed by these vectors.

The matrix formed by the coefficients of \( \hat{i}, \hat{j}, \hat{k} \) for vectors \( \mathbf{b}, \mathbf{c}, \mathbf{d} \) is:
\[ \begin{vmatrix} -1 & 2 & 1 \\ 1 & 1 & 1 \\ 2 & -1 & 0 \end{vmatrix} \]

### Step 3: Calculate the determinant
We can calculate the determinant using the formula for a 3x3 matrix:
\[ \text{det} = a(ei-fh) - b(di-fg) + c(dh-eg) \]
where the matrix is:
\[ \begin{vmatrix} a & b & c \\ d & e & f \\ g & h & i \end{vmatrix} \]

Substituting the values:
\[ \text{det} = -1 \cdot (1 \cdot 0 - 1 \cdot (-1)) - 2 \cdot (1 \cdot 0 - 1 \cdot 2) + 1 \cdot (1 \cdot (-1) - 2 \cdot 2) \]
Calculating each term:
1. First term: \( -1 \cdot (0 + 1) = -1 \)
2. Second term: \( -2 \cdot (0 - 2) = 4 \)
3. Third term: \( 1 \cdot (-1 - 4) = -5 \)

Adding these together:
\[ \text{det} = -1 + 4 - 5 = -2 \]

### Step 4: Analyze the determinant
Since the determinant is not equal to zero (\(-2 \neq 0\)), the vectors \( \mathbf{b}, \mathbf{c}, \mathbf{d} \) are non-coplanar.

### Step 5: Conclusion about Statement II
Since \( \mathbf{b}, \mathbf{c}, \mathbf{d} \) are non-coplanar, by Statement II, there exist real numbers \( \alpha, \beta, \gamma \) such that:
\[ \mathbf{a} = \alpha \mathbf{b} + \beta \mathbf{c} + \gamma \mathbf{d} \]
Thus, Statement II is true.

### Step 6: Conclusion about Statement I
However, Statement I claims that such \( \alpha, \beta, \gamma \) exist without confirming the coplanarity of \( \mathbf{b}, \mathbf{c}, \mathbf{d} \). Since we have established that they are non-coplanar, Statement I is also true.

### Final Answer
Both statements are true, and thus the answer is that both statements are valid.

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