The position vectors of three consecutive vertices of a parallelogram are$i + j + k, i + 3j + 5k and 7i + 9j + 11k.$ The position vector of the fourth vertex is

Let$A, B$ and $C$ be the three given vertices and $D$the fourth vertex, with the position vector $xi + yj + zk.$ Since $ABCD$ is a parallelogram, the diagonals $AC$ and $BD$ bisect each other so

$\begin{array}{r}\frac{(1+7)\mathbf{i}+(1+9)\mathbf{j}+(1+11)\mathbf{k}}{2}\\ =\frac{(1+x)\mathbf{i}+(3+y)\mathbf{j}+(5+z)\mathbf{k}}{2}\end{array}$

$\Rightarrow x=7,y=7\text{ and }z=7$ so that$D$ is the point 

$\Rightarrow$$7(\mathbf{i}+\mathbf{j}+\mathbf{k})$