The given solution of a second order linear differential equation is \(y=3e^{3x}\)

Evaluate y' as follows.

\(y'=\frac{d}{dx}(3e^{3x})\)

\(=3(e^{3x}\cdot3)\)

\(=9e^{3x}\)

Hence, \(y'=9e^{3x}\)

Evaluate y'' as follows.

\(y''=\frac{d}{dx}(9e^{3x})\)

\(=9(e^{3x}\cdot3)\)

\(=27e^{3x}.\)

So, \(y''=27e^{3x}.\)

Now, substitute the values of y' and y'' in all the options and verify.

Option (a):

\(y''-(3+a)y'+3ay=0\)

\((27e^{3x})-(3+a)(9e^{3x})+3a(3e^{3x})=0\)

\(27e^{3x}-27e^{3x}-9ae^{3x}+9ae^{3x}=0\)

\(0=0\)

Hence, option (a) satisfies the given solution.

Option (b):

\(y''+y'-6y=0\)

\(27e^{3x}+9e^{3x}-6(3e^{3x})=0\)

\(27^{3x}+9e^{3x}-18e^{3x}=0\)

\(18e^{3x}\neq0\)

Thus, option (b) does not satisfy the given solution.

Option (c):

\(y''+3y'=0\)

\(27e^{3x}+3(9e^{3x})=0\)

\(27e^{3x}+27e^{3x}=0\)

\(54e^{3x}\neq0\)

So, option (c) does not satisfy the given solution.

Option (d):

\(y''+(3-a)y'+3ay=0\)

\((27e^{3x})+(3-a)(9e^{3x})+3a(3e^{3x})=0\)

\(27e^{3x}+27e^{3x}-9ae^{3x}+9ae^{3x}=0\)

\(64e^{3x}\neq0\)

Hence, option (c) does not satisfy the given solution.

Since, the only option that satisfied is (a), the correct option is (a).

Therefore, the answer is option (a).

Evaluate y' as follows.

\(y'=\frac{d}{dx}(3e^{3x})\)

\(=3(e^{3x}\cdot3)\)

\(=9e^{3x}\)

Hence, \(y'=9e^{3x}\)

Evaluate y'' as follows.

\(y''=\frac{d}{dx}(9e^{3x})\)

\(=9(e^{3x}\cdot3)\)

\(=27e^{3x}.\)

So, \(y''=27e^{3x}.\)

Now, substitute the values of y' and y'' in all the options and verify.

Option (a):

\(y''-(3+a)y'+3ay=0\)

\((27e^{3x})-(3+a)(9e^{3x})+3a(3e^{3x})=0\)

\(27e^{3x}-27e^{3x}-9ae^{3x}+9ae^{3x}=0\)

\(0=0\)

Hence, option (a) satisfies the given solution.

Option (b):

\(y''+y'-6y=0\)

\(27e^{3x}+9e^{3x}-6(3e^{3x})=0\)

\(27^{3x}+9e^{3x}-18e^{3x}=0\)

\(18e^{3x}\neq0\)

Thus, option (b) does not satisfy the given solution.

Option (c):

\(y''+3y'=0\)

\(27e^{3x}+3(9e^{3x})=0\)

\(27e^{3x}+27e^{3x}=0\)

\(54e^{3x}\neq0\)

So, option (c) does not satisfy the given solution.

Option (d):

\(y''+(3-a)y'+3ay=0\)

\((27e^{3x})+(3-a)(9e^{3x})+3a(3e^{3x})=0\)

\(27e^{3x}+27e^{3x}-9ae^{3x}+9ae^{3x}=0\)

\(64e^{3x}\neq0\)

Hence, option (c) does not satisfy the given solution.

Since, the only option that satisfied is (a), the correct option is (a).

Therefore, the answer is option (a).