The point in the $xy$-plane which is equidistant from the points $A(2,0,3)$,$B(0,3,2)$,and $C(0,0,1)$ has the coordinates

$A(2,3,5), B(\alpha, 3,3)$ and $C(7,5, \beta)$ are the vertices of a triangle. If the median through $A$ is equally inclined with the coordinate axes,then $\cos^{-1}\left(\frac{\alpha}{\beta}\right) = $

Find the image of the point $(1, 2, 3)$ in the line $\vec{r} = (6\hat{i} + 7\hat{j} + 7\hat{k}) + \lambda(3\hat{i} + 2\hat{j} - 2\hat{k})$.

If $\left(\frac{9}{4}, \frac{5}{4}, \frac{15}{4}\right)$ is the centroid of a tetrahedron whose vertices are $(a, 2, 1), (1, b, 4), (4, 0, c)$ and $(1, 1, 7)$, then

If the direction cosines of two lines inclined at an angle $\theta$ are $(l_1, m_1, n_1)$ and $(l_2, m_2, n_2)$,then the direction cosines of the internal bisector of the angle between the lines are:

If a line makes angles $\alpha, \beta, \gamma, \delta$ with the four diagonals of a cube,then the value of $\cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma + \cos^2 \delta = $