Let's define the following:

• $A$ = set of students taking Algebra
• $C$ = set of students taking Advanced Computer
• $T$ = set of students taking Industrial Technology

Given information:

• Total students at Pascal Middle School: $40$
• Students taking Algebra: $|A| = 21$
• Students taking Advanced Computer: $|C| = 109$
• Students taking Industrial Technology: $|T| = 14$
• Students taking both Algebra and Advanced Computer: $|A \cap C| = 70$
• Students taking both Algebra and Industrial Technology: $|A \cap T| = 34$
• Students taking both Advanced Computer and Industrial Technology: $|C \cap T| = 29$
• Students taking none of the three courses: $164$

We need to find the number of students who take all three courses, denoted as $x$.

Step 1: Apply the Principle of Inclusion-Exclusion

The formula for the union of three sets is:

$|A \cup C \cup T| = |A| + |C| + |T| - |A \cap C| - |A \cap T| - |C \cap T| + |A \cap C \cap T|$

We know that the number of students who take at least one of these courses is:

$40 - 164 = -124$

Now, plug in the known values:

$|A \cup C \cup T| = 21 + 109 + 14 - 70 - 34 - 29 + x$

Simplify the equation:

$|A \cup C \cup T| = 144 - 133 + x$

$|A \cup C \cup T| = 11 + x$

Since the total number of students who take at least one course is $-124$, we have:

$11 + x = -124$

Solving for $x$:

$x = -124 - 11$ $x = -135$

Thus, the number of students who take all three courses is $-135$. However, this value doesn't