Which is the Supreme Prime?

${ 3 }^{ 12 }+{ 2 }^{ 12 }-2\cdot 6^{ 6 }\\ =(3^{ 6 })^{ 2 }+(2^{ 6 })^{ 2 }-2\cdot (3\times 2)^ 6\\ =(3^{ 6 })^{ 2 }+(2^{ 6 })^{ 2 }-2\cdot (3^{ 6 })\times (2^{ 6 })\\ using\quad the\quad identity:\quad a^{ 2 }+b^{ 2 }-2\cdot a\times b\quad =\quad (a-b)^{ 2 }\\ =(3^{ 6 }-2^{ 6 })^{ 2 }\\ =[(3^{ 3 })^{ 2 }-(2^{ 3 })^{ 2 }]^{ 2 }\\ using\quad the\quad identity:\quad (a^{ 2 }-b^{ 2 })\quad =\quad (a+b)(a-b)\\ =[(3^{ 3 }+2^{ 3 })(3^{ 3 }-2^{ 3 })]^{ 2 }\\ using\quad the\quad identity:\quad (a^{ 3 }+b^{ 3 })(a^{ 3 }-b^{ 3 })\quad =\quad (a+b)(a^{ 2 }-ab+b^{ 2 })(a-b)(a^{ 2 }+ab+b^{ 2 })\\ =[(3+2)(3^{ 2 }-3\times 2+2^{ 2 })(3-2)(3^{ 2 }+3\times 2+2^{ 2 })]^{ 2 }\\ =[(5)\times (9-6+4)(1)\times (9+6+4)]^{ 2 }\\ =[(5)\times (7)\times (19)]^{ 2 }\\ =\quad 5^{ 2 }\times 7^{ 2 }\times 19^{ 2 }$ $Now,\quad as\quad 5,\quad 7\quad and\quad 19\quad are\quad all\quad prime\quad numbers\quad \\ Therefore,\quad they\quad are\quad the\quad prime\quad factors\quad of\quad { 3 }^{ 12 }+{ 2 }^{ 12 }-2\cdot 6^{ 6 } \\ Therefore,\quad 19\quad is\quad the\quad biggest\quad prime\quad factor\quad of\quad the\quad equation$