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.