The centroid of a triangle with vertices $A(3,4,5)$, $B(6,7,2)$, and $C(x, y, z)$ is $(3,2,3)$. Then $x+y+z=$

Match the following columns:

Column $I$	Column $II$
$(A)$ The centroid of the triangle formed by $(2, 3, -1)$,$(5, 6, 3)$,$(2, -3, 1)$ is	$(p)$ $(2, 2, 2)$
$(B)$ The circumcentre of the triangle formed by $(1, 2, 3)$,$(2, 3, 1)$,$(3, 1, 2)$ is	$(q)$ $(3, 1, 4)$
$(C)$ The orthocentre of the triangle formed by $(2, 1, 5)$,$(3, 2, 3)$,$(4, 0, 4)$ is	$(r)$ $(1, 1, 0)$
$(D)$ The incentre of the triangle formed by $(0, 0, 0)$,$(3, 0, 0)$,$(0, 4, 0)$ is	$(s)$ $(3, 2, 1)$

If the points $P = \hat{i} + 2 \hat{j}$,$Q = 4 \hat{i} + 6 \hat{j}$,$R = 5 \hat{i} + 7 \hat{j}$,and $S = a \hat{i} + b \hat{j}$ are the consecutive vertices of a parallelogram $PQRS$,then:

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

Consider a cuboid where all edges are integers and the base is a square. Suppose the sum of all its edges is numerically equal to the sum of the areas of all its six faces. Then,the sum of all its edges is

Let $ABCD$ be a parallelogram and $E$ be the mid-point of $AB$. If $P$ is the point of intersection of $DE$ and $AC$,then $\frac{DP}{PE} + \frac{AP}{PC} = $