If a line makes angles $\alpha ,\beta ,\gamma$ with the positive directions of x-axis, y-axis and z-axis respectively, then ${\sin }^2\alpha +{\sin }^2\beta +{\sin }^2\gamma$ is equal to

The correct answer is 2.

To understand why, let us recall the concept of direction cosines of a line in three-dimensional space.
If a line makes angles $\alpha$, $\beta$, and $\gamma$ with the positive directions of the x-axis, y-axis, and z-axis respectively, then the direction cosines of the line are defined as:
$l=\cos \alpha$
$m=\cos \beta$
$n=\cos \gamma$

A fundamental property of direction cosines is that the sum of their squares is always equal to 1:
$l^2+m^2+n^2=1$
Substituting the values of $l$, $m$, and $n$, we get:
${\cos }^2\alpha +{\cos }^2\beta +{\cos }^2\gamma =1$

We want to find the value of ${\sin }^2\alpha +{\sin }^2\beta +{\sin }^2\gamma$.
Using the basic trigonometric identity ${\sin }^2\theta +{\cos }^2\theta =1$, we can write:
${\cos }^2\alpha =1-{\sin }^2\alpha$
${\cos }^2\beta =1-{\sin }^2\beta$
${\cos }^2\gamma =1-{\sin }^2\gamma$

Substituting these expressions into our relation, we get:
$(1-{\sin }^2\alpha )+(1-{\sin }^2\beta )+(1-{\sin }^2\gamma )=1$

Simplifying the left-hand side:
$3-({\sin }^2\alpha +{\sin }^2\beta +{\sin }^2\gamma )=1$

Rearranging the terms to solve for the target expression:
${\sin }^2\alpha +{\sin }^2\beta +{\sin }^2\gamma =3-1$
${\sin }^2\alpha +{\sin }^2\beta +{\sin }^2\gamma =2$

Thus, the expression is equal to 2.