vector `vecc` are perpendicular to vectors `veca= (2,-3,1) and vecb= (1, -2,3) ` and satifies the condition `vecc. (hati + 2hatj - 7 hatk) = 10` then vector `vecc` is equal to
`(a) (7,5,1)` `(b) (-7,-5,-1)` `(c) (1,1,-1)` `(d)` none of these

Options

A. 7,5,1
B. (-7, -5, -1)
C. 1,1,-1
D. none of these

Explanation

To find the vector \(\vec{c}\) that is perpendicular to both vectors \(\vec{a} = (2, -3, 1)\) and \(\vec{b} = (1, -2, 3)\), and satisfies the condition \(\vec{c} \cdot (\hat{i} + 2\hat{j} - 7\hat{k}) = 10\), we can follow these steps:

### Step 1: Find the cross product \(\vec{a} \times \vec{b}\)

The cross product of two vectors \(\vec{a}\) and \(\vec{b}\) can be calculated using the determinant of a matrix formed by the unit vectors and the components of the vectors.

\[ \vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & -3 & 1 \\ 1 & -2 & 3 \end{vmatrix} \]

Calculating the determinant:

\[ = \hat{i} \begin{vmatrix} -3 & 1 \\ -2 & 3 \end{vmatrix} - \hat{j} \begin{vmatrix} 2 & 1 \\ 1 & 3 \end{vmatrix} + \hat{k} \begin{vmatrix} 2 & -3 \\ 1 & -2 \end{vmatrix} \]

Calculating each of the 2x2 determinants:

1. \(\begin{vmatrix} -3 & 1 \\ -2 & 3 \end{vmatrix} = (-3)(3) - (1)(-2) = -9 + 2 = -7\)

2. \(\begin{vmatrix} 2 & 1 \\ 1 & 3 \end{vmatrix} = (2)(3) - (1)(1) = 6 - 1 = 5\)

3. \(\begin{vmatrix} 2 & -3 \\ 1 & -2 \end{vmatrix} = (2)(-2) - (-3)(1) = -4 + 3 = -1\)

Putting these values back into the cross product:

\[ \vec{a} \times \vec{b} = -7\hat{i} - 5\hat{j} - 1\hat{k} \]

### Step 2: Express \(\vec{c}\) in terms of \(\lambda\)

Since \(\vec{c}\) is perpendicular to both \(\vec{a}\) and \(\vec{b}\), we can express \(\vec{c}\) as:

\[ \vec{c} = \lambda(-7\hat{i} - 5\hat{j} - \hat{k}) \]

### Step 3: Use the condition \(\vec{c} \cdot (\hat{i} + 2\hat{j} - 7\hat{k}) = 10\)

Substituting \(\vec{c}\) into the dot product condition:

\[ \lambda(-7\hat{i} - 5\hat{j} - \hat{k}) \cdot (\hat{i} + 2\hat{j} - 7\hat{k}) = 10 \]

Calculating the dot product:

\[ = \lambda \left[ -7(1) + (-5)(2) + (-1)(-7) \right] \]
\[ = \lambda \left[ -7 - 10 + 7 \right] \]
\[ = \lambda(-10) \]

Setting this equal to 10:

\[ -10\lambda = 10 \implies \lambda = -1 \]

### Step 4: Find \(\vec{c}\)

Substituting \(\lambda\) back into the expression for \(\vec{c}\):

\[ \vec{c} = -1(-7\hat{i} - 5\hat{j} - \hat{k}) = (7\hat{i} + 5\hat{j} + \hat{k}) \]

Thus, the vector \(\vec{c}\) is:

\[ \vec{c} = (7, 5, 1) \]

### Conclusion

The correct answer is (a) \((7, 5, 1)\).