As shown in the figure above, ABCD is a parallelogram which has sides of length 15 and 10 .

Now, we can check from the options that which of them hold true

(A) The integral value of the length of the diagonal $A C$ is at least 19 if it is the longer diagonal in the parallelogram -

Since triangle ABC is an obtuse angled triangle, where angle ABC is obtuse, we get $\(\mathrm{AC}^2>\mathrm{AB}^2+\mathrm{BC}^2\)$ i.e. $\(\mathrm{AC}^2>15^2+10^2=325 \Rightarrow \mathrm{AC}>\sqrt{325}>18\)$. Thus, the least integral value of $A C$ is 19 , so the statement $(A)$ is true.

(B) The integral value of the length of the diagonal $B D$ is at most 18 , if it is the shorter diagonal in the parallelogram -

Since triangle ABD is an acute triangle, we get $\(\mathrm{BD}^2<\mathrm{AD}^2+\mathrm{AB}^2\)$ i.e. $\(\mathrm{BD}^2<15^2+10^2=325\)$, which gives $\(\mathrm{BD}<\sqrt{325}<19\)$. Thus, the maximum integral value of AC is 18 , so the statement (B) is true.

(C) The integral value of the area of the parallelogram cannot exceed 150 sq. units Area of parallelogram ABCD is $\(=2\left(\frac{1}{2} \times \mathrm{AB} \times \mathrm{AD} \times \sin \angle \mathrm{BAD}\right)=\mathrm{AB} \times \mathrm{AD} \times \sin \angle \mathrm{BAD}\)$ which is same as $\(15 \times 10 \times \sin \angle \mathrm{BAD}\)$

The area would be maximum when $\(\sin \angle \mathrm{BAD}\)$ is maximum which would happen when $\(\angle \mathrm{BAD}=90^{\circ}\)$, which gives $\(\sin \angle \mathrm{BAD}=\sin 90^{\circ}=1\)$. So, the maximum possible area is $\(15 \times 10 \times 1=150\)$

Thus the area of the parallelogram $A B C D$ cannot exceed 150 , so statement $(C)$ is true.

Hence options (A), (B) \& (C) are correct.