Let a unit vector `vec OP` make angles `alpha,beta,gamma` with the positive directions of the co-ordinate axes `OX,OY,OZ` respectively,where `beta in(0,(pi)/(2))`.If `vec OP` is perpendicular to the plane through points `(1,2,3),(2,3,4)` and `(1,5,7),` then which one of the following is true?

Options

A. `alpha in(0,(pi)/(2))` and `gamma in(0,(pi)/(2))
B. `alpha in(pi/2,pi)` and `gamma in((pi)/(2),pi)`
C. `alpha in((pi)/(2),pi)` and `gamma in(0,(pi)/(2))`
D. `alpha in(0,(pi)/(2))` and `gamma in((pi)/(2),pi)`

Answer

To solve the problem, we need to find the unit vector \(\vec{OP}\) that is perpendicular to the plane formed by the points \(A(1,2,3)\), \(B(2,3,4)\), and \(C(1,5,7)\).

### Step 1: Find the vectors \(\vec{AB}\) and \(\vec{AC}\)
First, we calculate the vectors \(\vec{AB}\) and \(\vec{AC}\):

\[ \vec{AB} = B - A = (2-1, 3-2, 4-3) = (1, 1, 1) \]
\[ \vec{AC} = C - A = (1-1, 5-2, 7-3) = (0, 3, 4) \]

### Step 2: Calculate the normal vector \(\vec{n}\) using the cross product
Next, we find the normal vector \(\vec{n}\) to the plane by taking the cross product of \(\vec{AB}\) and \(\vec{AC}\):

\[ \vec{n} = \vec{AB} \times \vec{AC} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 1 \\ 0 & 3 & 4 \end{vmatrix} \]

Calculating the determinant, we have:

\[ \vec{n} = \hat{i} \begin{vmatrix} 1 & 1 \\ 3 & 4 \end{vmatrix} - \hat{j} \begin{vmatrix} 1 & 1 \\ 0 & 4 \end{vmatrix} + \hat{k} \begin{vmatrix} 1 & 1 \\ 0 & 3 \end{vmatrix} \]

Calculating each of the determinants:

- For \(\hat{i}\): \(1 \cdot 4 - 1 \cdot 3 = 4 - 3 = 1\)
- For \(\hat{j}\): \(1 \cdot 4 - 1 \cdot 0 = 4\)
- For \(\hat{k}\): \(1 \cdot 3 - 1 \cdot 0 = 3\)

Thus,

\[ \vec{n} = \hat{i}(1) - \hat{j}(4) + \hat{k}(3) = (1, -4, 3) \]

### Step 3: Normalize the normal vector to get the unit vector \(\vec{OP}\)
To find the unit vector \(\vec{OP}\), we normalize \(\vec{n}\):

\[ |\vec{n}| = \sqrt{1^2 + (-4)^2 + 3^2} = \sqrt{1 + 16 + 9} = \sqrt{26} \]

The unit vector \(\vec{OP}\) is given by:

\[ \vec{OP} = \frac{1}{\sqrt{26}}(1, -4, 3) = \left(\frac{1}{\sqrt{26}}, -\frac{4}{\sqrt{26}}, \frac{3}{\sqrt{26}}\right) \]

### Step 4: Determine the angles \(\alpha\), \(\beta\), and \(\gamma\)
The angles \(\alpha\), \(\beta\), and \(\gamma\) that the unit vector makes with the coordinate axes can be found using the cosine formula:

\[ \cos(\alpha) = \frac{1}{\sqrt{26}}, \quad \cos(\beta) = -\frac{4}{\sqrt{26}}, \quad \cos(\gamma) = \frac{3}{\sqrt{26}} \]

### Step 5: Analyze the angle \(\beta\)
Given that \(\beta \in (0, \frac{\pi}{2})\), we have:

- \(\cos(\beta) = -\frac{4}{\sqrt{26}}\) implies \(\beta\) is in the second quadrant, which contradicts the condition that \(\beta\) is in \((0, \frac{\pi}{2})\).

### Conclusion
Since \(\beta\) must be between \(0\) and \(\frac{\pi}{2}\), we can conclude that the angles \(\alpha\) and \(\gamma\) must be in the range \(\left[\frac{\pi}{2}, \pi\right)\).

Thus, the correct option is that \(\alpha\) and \(\gamma\) are in the range \(\left[\frac{\pi}{2}, \pi\right)\).