A line makes angles α , β and γ with the positive directions of the...

Let $\cos\alpha$, $\cos\beta$, and $\cos\gamma$ be the direction cosines of the given line.

We know that the sum of the squares of the direction cosines is $1$:

$\cos^2\alpha + \cos^2\beta + \cos^2\gamma = 1$

The given vectors are:

$\vec{a} = (\sin^2\alpha)\hat{i} + (\sin^2\beta)\hat{j} + (\sin^2\gamma)\hat{k}$

$\vec{b} = \hat{i} + \hat{j} + \hat{k}$

The dot product $\vec{a} \cdot \vec{b}$ is given by:

$\vec{a} \cdot \vec{b} = (\sin^2\alpha)(1) + (\sin^2\beta)(1) + (\sin^2\gamma)(1)$

$\vec{a} \cdot \vec{b} = \sin^2\alpha + \sin^2\beta + \sin^2\gamma$

Using the identity $\sin^2\theta = 1 - \cos^2\theta$, we get:

$\vec{a} \cdot \vec{b} = (1 - \cos^2\alpha) + (1 - \cos^2\beta) + (1 - \cos^2\gamma)$

$\vec{a} \cdot \vec{b} = 3 - (\cos^2\alpha + \cos^2\beta + \cos^2\gamma)$

Substituting $\cos^2\alpha + \cos^2\beta + \cos^2\gamma = 1$:

$\vec{a} \cdot \vec{b} = 3 - 1 = 2$

Answer: $2$