The vectors `hati+2hatj+3hatk,lamdahati+4hatj+7hatk,-3hati-2hatj-5hatk` are collinear, of `lamda` is equal to (A)3 (B)4 (C)5 (D)6

Options

A. 3
B. 4
C. 5
D. 6

Explanation

To determine the value of λ for which the vectors \( \hat{i} + 2\hat{j} + 3\hat{k} \), \( \lambda \hat{i} + 4\hat{j} + 7\hat{k} \), and \( -3\hat{i} - 2\hat{j} - 5\hat{k} \) are collinear, we can set up a determinant of the vectors and solve for λ.

### Step-by-Step Solution:

1. **Write the vectors in matrix form**:
The vectors can be represented as rows in a matrix:
\[ \begin{vmatrix} 1 & 2 & 3 \\ \lambda & 4 & 7 \\ -3 & -2 & -5 \end{vmatrix} \]

2. **Set the determinant equal to zero**:
For the vectors to be collinear, the determinant of this matrix must equal zero:
\[ \begin{vmatrix} 1 & 2 & 3 \\ \lambda & 4 & 7 \\ -3 & -2 & -5 \end{vmatrix} = 0 \]

3. **Calculate the determinant**:
We can calculate the determinant using the formula for a 3x3 matrix:
\[ D = a(ei - fh) - b(di - fg) + c(dh - eg) \]
where \( a, b, c \) are the elements of the first row, and \( d, e, f, g, h, i \) are the elements of the second and third rows respectively.

Substituting the values:
\[ D = 1 \cdot (4 \cdot (-5) - 7 \cdot (-2)) - 2 \cdot (\lambda \cdot (-5) - 7 \cdot (-3)) + 3 \cdot (\lambda \cdot (-2) - 4 \cdot (-3)) \]

Simplifying each part:
- First term: \( 1 \cdot (-20 + 14) = 1 \cdot (-6) = -6 \)
- Second term: \( -2 \cdot (-5\lambda + 21) = 10\lambda - 42 \)
- Third term: \( 3 \cdot (-2\lambda + 12) = -6\lambda + 36 \)

Putting it all together:
\[ D = -6 + 10\lambda - 42 - 6\lambda + 36 \]
Combine like terms:
\[ D = (10\lambda - 6\lambda) + (-6 - 42 + 36) = 4\lambda - 12 \]

4. **Set the determinant to zero**:
Now, we set the determinant equal to zero:
\[ 4\lambda - 12 = 0 \]

5. **Solve for λ**:
\[ 4\lambda = 12 \implies \lambda = \frac{12}{4} = 3 \]

### Conclusion:
The value of \( \lambda \) for which the vectors are collinear is \( \lambda = 3 \).