$\text{ Two sides of a parallelogram are along the lines }\mathrm{x}+\mathrm{y}=3\text{ and }\mathrm{x}-\mathrm{y}+3=0\text{ . If its }$

$\text{ diagonals intersect at }(2,4)\text{ then one of its vertex is }$

 solving the given equations A(0,3) and  $P\left(2,4\right)$ is the midpoint 

using midpoint formula we get C(4,5)

given lines are perpendicular hence the figure formed is a rectangle

let the equation of BC be x-y+k=0 sub C we get k=1

let the equation of CD be x+y+c=0 sub C we get c=-9

solving BC and AB we get B(1,2)

solving CD and AD we get D(3,6)

$D\left(3,6\right)$