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

Let the three given points be $$A(1,2,3),\;B(2,3,4),\;C(1,5,7)$$. Any vector perpendicular to the plane $$ABC$$ is parallel to $$\overrightarrow{AB}\times\overrightarrow{AC}$$.

First write the two vectors lying in the plane: $$\overrightarrow{AB}=B-A=(1,1,1),\qquad \overrightarrow{AC}=C-A=(0,3,4).$$

Using the formula for a cross-product, $$\overrightarrow{AB}\times\overrightarrow{AC}= \begin{vmatrix} \hat i & \hat j & \hat k\\ 1 & 1 & 1\\ 0 & 3 & 4 \end{vmatrix} = \hat i(1\cdot4-1\cdot3)-\hat j(1\cdot4-1\cdot0)+\hat k(1\cdot3-1\cdot0) = 1\hat i-4\hat j+3\hat k.$$

Thus one normal vector is $$(1,-4,3)$$. Any scalar multiple of it is also normal to the plane. Because the required vector $$\vec{OP}$$ must be a unit vector and must satisfy $$\beta\in(0,\tfrac{\pi}{2})$$ (i.e. its y-component positive), we choose the opposite direction: $$(-1,4,-3).$$ This has a positive y-component ($$+4$$), so $$\beta$$ indeed lies between $$0$$ and $$\tfrac{\pi}{2}$$.

Convert this direction to a unit vector: $$\sqrt{(-1)^2+4^2+(-3)^2}=\sqrt{1+16+9}=\sqrt{26},$$ so $$\vec{OP}= \frac{-1}{\sqrt{26}}\hat i+\frac{4}{\sqrt{26}}\hat j+\frac{-3}{\sqrt{26}}\hat k.$$

Therefore $$\cos\alpha=\frac{-1}{\sqrt{26}}\lt0,\qquad \cos\beta=\frac{4}{\sqrt{26}}\gt0,\qquad \cos\gamma=\frac{-3}{\sqrt{26}}\lt0.$$

When the cosine is negative, the angle lies in $$(\tfrac{\pi}{2},\pi)$$; when the cosine is positive, the angle lies in $$(0,\tfrac{\pi}{2})$$. Hence $$\alpha\in\left(\tfrac{\pi}{2},\pi\right),\qquad \beta\in\left(0,\tfrac{\pi}{2}\right),\qquad \gamma\in\left(\tfrac{\pi}{2},\pi\right).$$

Comparing with the given options, this matches Option A: $$\alpha \in (\tfrac{\pi}{2},\pi)\ \text{and}\ \gamma \in (\tfrac{\pi}{2},\pi).$$