Find the projection of the line segment joining the points $A(-1, 2, 3)$ and $B(-1, 4, 0)$ on a line which makes angles of $45^{\circ}, 60^{\circ},$ and $60^{\circ}$ with the coordinate axes.

The direction cosines of the line $\frac{3x + 1}{-3} = \frac{3y + 2}{6} = \frac{z}{-1}$ are

The distance between two points $P$ and $Q$ is $d$ and the lengths of the projections of $PQ$ on the coordinate planes are $d_1, d_2, d_3$. Then $d_1^2 + d_2^2 + d_3^2 = k d^2$,where $k$ is:

If the projections of a line segment on the coordinate axes are $2, 3, 6$,find the length of the line segment.

$A$ space vector makes the angles $150^{\circ}$ and $60^{\circ}$ with the positive direction of $x$- and $y$-axes. The angle made by the vector with the positive direction of the $z$-axis is (in $^{\circ}$)

Suppose $(l_1, m_1, n_1)$ and $(l_2, m_2, n_2)$ are the direction cosines of two lines and $\theta$ is the angle between them,where $\cos \theta = \pm(l_1 l_2 + m_1 m_2 + n_1 n_2)$. Let $A=(1, -2, 3)$,$B=(3, 1, -3)$,and $C=(-3, 1, 3)$ be the vertices of $\triangle ABC$. Then,$\cos A =$