we will proceed to verify each case to determine the solution
remember that
[tex]i^{2} =-1[/tex]
case A) [tex](1 + 3i)(6i)[/tex]
applying distributive property
[tex](1 + 3i)(6i)=1*6i+3i*6i \\=6i+18i^{2}\\=6i+18*(-1)\\=6i-18[/tex]
[tex]-18+6i[/tex] ------> is not a real number
therefore
the case A) is not a real number product
case B) [tex](1 + 3i)(2-3i)[/tex]
applying distributive property
[tex](1 + 3i)(2-3i)=1*2+1*(-3i)+3i*(2)+3i*(-3i)\\=2-3i+6i-9i^{2}\\=2+3i-9*(-1) \\=11+3i[/tex]
[tex]11+3i[/tex] ------> is not a real number
therefore
the case B) is not a real number product
case C) [tex](1 + 3i)(1-3i)[/tex]
applying difference of square
[tex](1 + 3i)(1-3i)=(1)^{2}-(3i)^{2}\\=1-9i^{2}\\=1-9*(-1) \\=10[/tex]
[tex]10[/tex] ------> is a real number
therefore
the case C) is a real number product
case D) [tex](1 + 3i)(3i)[/tex]
applying distributive property
[tex](1 + 3i)(3i)=1*3i+3i*3i \\=3i+9i^{2}\\=3i+9*(-1) \\=3i-9[/tex]
[tex]-9+3i[/tex] ------> is not a real number
therefore
the case D) is not a real number product
the answer is
[tex](1 + 3i)(1-3i)[/tex]
