The points $(5, -4, 2), (4, -3, 1), (7, -6, 4)$ and $(8, -7, 5)$ are the vertices of

If the points $(-1, 3, 2)$,$(-4, 2, -2)$,and $(5, 5, \lambda)$ are collinear,then $\lambda = $

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 $\frac{9 \sqrt{2}}{2}$.

If $A(1, 2, -3)$,$B(2, 3, -1)$,and $C(3, 1, 1)$ are the vertices of $\triangle ABC$,then $\left|\frac{\cos A}{\cos B}\right| = $

$G(1,0,1)$ is the centroid of the triangle $ABC$. If $A=(1,-4,2)$ and $B=(3,1,0)$,then $AG^2+CG^2=$

$A(2, 3, k)$, $B(-1, k, -1)$, and $C(4, -3, 2)$ are the vertices of $\triangle ABC$. If $AB = AC$ and $k > 0$, then $\triangle ABC$ is: