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To solve for \( t \) given that \( \mathbf{a} = t\mathbf{i} + 2\mathbf{j} + 3\mathbf{k} \) and \( |\mathbf{a}| = 7 \), we start by finding the magnitude of \( \mathbf{a} \).

1. The magnitude of \( \mathbf{a} \) is given by:
   \[
   |\mathbf{a}| = \sqrt{(t)^2 + (2)^2 + (3)^2}
   \]

2. Substitute the values for each component:
   \[
   |\mathbf{a}| = \sqrt{t^2 + 4 + 9}
   \]

3. Simplify inside the square root:
   \[
   |\mathbf{a}| = \sqrt{t^2 + 13}
   \]

4. Since \( |\mathbf{a}| = 7 \), we set up the equation:
   \[
   \sqrt{t^2 + 13} = 7
   \]

5. Square both sides to remove the square root:
   \[
   t^2 + 13 = 49
   \]

6. Subtract 13 from both sides:
   \[
   t^2 = 36
   \]

7. Take the square root of both sides to solve for \( t \):
   \[
   t = \pm 6
   \]

### Final Answer:
The possible values of \( t \) are \( t = 6 \) and \( t = -6 \).

Would you like a further breakdown of any steps or more details?

Here are some related questions that might deepen your understanding:
1. How is the magnitude of a vector calculated in 3D space?
2. What are the implications of having multiple solutions when taking square roots?
3. How does the Pythagorean theorem relate to calculating the magnitude of vectors?
4. Why is it necessary to square both sides of the equation when dealing with square roots?
5. How can we verify the solutions obtained for \( t \) in this problem?

**Tip:** When dealing with vector magnitudes, always remember that taking the square root can yield both positive and negative solutions for the variable involved.

Given that a = ti + 2j + 3k, and that |a| = 7, find the possible values of t.

Explore the solution to finding the possible values of t for the vector a = ti + 2j + 3k, where the magnitude |a| is 7. This problem involves vector magnitudes, algebra, and solving quadratic equations, suitable for students in Grades 10-12.

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