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

If the origin is the centroid of a triangle $ABC$ having vertices $A(a, 1, 3)$,$B(-2, b, -5)$,and $C(4, 7, c)$,find the values of $a, b, c$.

If the points $(x, -1)$,$(3, y)$,$(-2, 3)$,and $(-3, -2)$ are the vertices of a parallelogram,then:

Let $A(x, y, z)$ be a point in the $xy$-plane,which is equidistant from three points $P(0, 3, 2)$,$Q(2, 0, 3)$,and $R(0, 0, 1)$. Let $B = (1, 4, -1)$ and $C = (2, 0, -2)$. Then among the statements $(S1) :$ $\triangle ABC$ is an isosceles right-angled triangle and $(S2) :$ the area of $\triangle ABC$ is $\frac{9 \sqrt{2}}{2}$.

$\triangle ABC$ has vertices at $A \equiv (2, 3, 5)$,$B \equiv (-1, 3, 2)$ and $C \equiv (\lambda, 5, \mu)$. If the median through $A$ is equally inclined to the axes,then the values of $\lambda$ and $\mu$ respectively are

In $\triangle ABC$,if the midpoints of the sides $AB, BC$ and $CA$ are respectively $(l, 0, 0), (0, m, 0)$ and $(0, 0, n)$,then $\frac{AB^2+BC^2+CA^2}{l^2+m^2+n^2}=$