SQLite - INSERT INTO SELECT - how to insert data of "join of 3 existing tables into a new table"?

So the scenario here is, I have 4 tables in the database namely:

1. 
   

   "question_info":
   
   CREATE TABLE question_info ( q_id mediumint(9) NOT NULL, q_type_id int(11) NOT NULL, q_options_id mediumint(9) NOT NULL, q_category_id int(11) NOT NULL, q_text varchar(2048) NOT NULL, status tinyint(4) NOT NULL DEFAULT '0', q_date_added date NOT NULL DEFAULT '2013-01-01', q_difficulty_level tinyint(4) NOT NULL DEFAULT '0', PRIMARY KEY(q_id) );
   

   
   


2. 
   

   "question_options_info":
   
   CREATE TABLE question_options_info ( q_options_id mediumint(9) NOT NULL, q_options_1 varchar(255) NOT NULL, q_options_2 varchar(255) NOT NULL, q_options_3 varchar(255) NOT NULL, q_options_4 varchar(255) NOT NULL, q_options_ex_1 varchar(1024) DEFAULT NULL, q_options_ex_2 varchar(1024) DEFAULT NULL, q_options_ex_3 varchar(1024) DEFAULT NULL, q_options_ex_4 varchar(1024) DEFAULT NULL, PRIMARY KEY(q_options_id) );

   
   


3. 
   

   "question_answer_info":
   
   CREATE TABLE question_answer_info ( q_id mediumint(9) NOT NULL, q_options mediumint(9) NOT NULL );

   
   


4. 
   

   "trivia_data":
   
   CREATE TABLE trivia_data ( q_id mediumint(9) NOT NULL, q_text varchar(2048) NOT NULL, q_options_1 varchar(255) NOT NULL, q_options_2 varchar(255) NOT NULL, q_options_3 varchar(255) NOT NULL, q_options_4 varchar(255) NOT NULL, q_options mediumint(9) NOT NULL, q_difficulty_level tinyint(4) NOT NULL DEFAULT '0', q_date_added date NOT NULL DEFAULT '2015-04-8', PRIMARY KEY(q_id) );

   
   

So what I need is to, insert a data into trivia_data table. The data is returned by this query:



SELECT question_info.q_id, question_info.q_text, question_options_info.q_options_1, question_options_info.q_options_2, question_options_info.q_options_3, question_options_info.q_options_4, question_answer_info.q_options, question_info.q_difficulty_level, question_info.q_date_added

FROM question_info JOIN question_options_info ON question_info.q_options_id = question_options_info.q_options_id JOIN question_answer_info ON question_info.q_id = question_answer_info.q_id;

This query would return data somewhat like this:

I have already tried this specific query to insert the data:

INSERT INTO trivia_data VALUES(q_id, q_text, q_options_1, q_options_2, q_options_3, q_options_4, q_options, q_difficulty_level, q_date_added) SELECT question_info.q_id, question_info.q_text, question_options_info.q_options_1, question_options_info.q_options_2, question_options_info.q_options_3, question_options_info.q_options_4, question_answer_info.q_options, question_info.q_difficulty_level, question_info.q_date_added FROM question_info JOIN question_options_info on question_info.q_options_id = question_options_info.q_options_id JOIN question_answer_info on question_info.q_id = question_answer_info.q_id;

But it always returns this error:

near "SELECT": syntax error:

Honestly I am a novice to SQL. So please try to explain as simply as possible. Any help would be appreciated. Thank You.