Which of the following statements is incorrect?

Correct option is D

(a) Given a CFG G and a string x, an algorithm can decide whether x ∈ L(G) in polynomial time. — True
The membership problem for Context-Free Languages (CFLs) is decidable in polynomial time.
For example, the CYK (Cocke–Younger–Kasami) algorithm can determine whether a string belongs to the language generated by a CFG in $$O(n^3)$$​ time for a grammar in Chomsky Normal Form.
(b) Given a context-free language L, its complement is also context free. — False
Context-Free Languages are not closed under complementation.
That is:
​$$L\text{ is CFL}\not\Rightarrow\overline L\text{ is CFL}$$​​
For example, consider:
​$$L=\{a^ib^jc^k\mid i,j,k\geq0\text{ and }i=j=k\}$$​​
The language
​$$\{a^nb^nc^n\mid n\geq0\}$$​​
is not context-free. Using closure properties, one can construct examples demonstrating that CFLs are not closed under complement.
(c) $$\{a^n b^m c^p:m,n,p\geq0,\ m\ne n\text{ or }m\ne p\}$$​ is not deterministic context-free. — True
Let:
​$$L=\{a^n b^m c^p:m\ne n\text{ or }m\ne p\}$$​​
Within the regular language:
​$$R=a^*b^*c^*$$​​
the complement of L is:
​$$R-L=\{a^n b^n c^n:n\geq0\}$$​​
Now, suppose L were a deterministic context-free language (DCFL).
DCFLs are closed under complement, and regular languages are also DCFLs. Therefore, R – L would also be a DCFL and hence a CFL.
But, $$\{a^nb^nc^n:n\geq0\}$$​ is not context-free.
This gives a contradiction.
(d) Class of deterministic context-free languages is closed under intersection. — False
DCFLs are not closed under intersection.
That is:
​$$L_1,L_2\text{ are DCFLs}$$​​
does not necessarily imply:
$$L_1\cap L_2$$​ is a DCFL.
However, DCFLs are closed under complement.