To find the length of the perpendicular from point \(P\) to the line \(l\), we can use the vector projection method.

1. The direction vector of the line \(l\) is \(\mathbf{d} = \mathbf{i} + 2\mathbf{j} + 3\mathbf{k}\).

2. The position vector of point \(P\) is \(\mathbf{p} = 3\mathbf{i} - 2\mathbf{j} + \mathbf{k}\).

3. A point \(A\) on the line \(l\) can be represented as \(\mathbf{a} = 4\mathbf{i} + 2\mathbf{j} + 5\mathbf{k}\).

4. The vector \(\mathbf{PA}\) is \(\mathbf{p} - \mathbf{a} = (3 - 4)\mathbf{i} + (-2 - 2)\mathbf{j} + (1 - 5)\mathbf{k} = -\mathbf{i} - 4\mathbf{j} - 4\mathbf{k}\).

5. The projection of \(\mathbf{PA}\) onto \(\mathbf{d}\) is given by:

\(\text{proj}_{\mathbf{d}} \mathbf{PA} = \frac{\mathbf{PA} \cdot \mathbf{d}}{\mathbf{d} \cdot \mathbf{d}} \mathbf{d}\)

6. Calculate \(\mathbf{PA} \cdot \mathbf{d} = (-1)(1) + (-4)(2) + (-4)(3) = -1 - 8 - 12 = -21\).

7. Calculate \(\mathbf{d} \cdot \mathbf{d} = 1^2 + 2^2 + 3^2 = 1 + 4 + 9 = 14\).

8. The projection is:

\(\text{proj}_{\mathbf{d}} \mathbf{PA} = \frac{-21}{14} \mathbf{d} = -\frac{3}{2} \mathbf{d}\)

9. The perpendicular vector \(\mathbf{v}\) from \(P\) to \(l\) is:

\(\mathbf{v} = \mathbf{PA} - \text{proj}_{\mathbf{d}} \mathbf{PA}\)

10. Calculate \(\mathbf{v} = (-\mathbf{i} - 4\mathbf{j} - 4\mathbf{k}) - \left(-\frac{3}{2}(\mathbf{i} + 2\mathbf{j} + 3\mathbf{k})\right)\).

11. Simplify \(\mathbf{v} = -\mathbf{i} - 4\mathbf{j} - 4\mathbf{k} + \frac{3}{2}\mathbf{i} + 3\mathbf{j} + \frac{9}{2}\mathbf{k}\).

12. \(\mathbf{v} = \left(-1 + \frac{3}{2}\right)\mathbf{i} + (-4 + 3)\mathbf{j} + \left(-4 + \frac{9}{2}\right)\mathbf{k}\).

13. \(\mathbf{v} = \frac{1}{2}\mathbf{i} - \mathbf{j} + \frac{1}{2}\mathbf{k}\).

14. The length of \(\mathbf{v}\) is:

\(\|\mathbf{v}\| = \sqrt{\left(\frac{1}{2}\right)^2 + (-1)^2 + \left(\frac{1}{2}\right)^2} = \sqrt{\frac{1}{4} + 1 + \frac{1}{4}} = \sqrt{\frac{3}{2}}\)

15. \(\|\mathbf{v}\| \approx 1.22\) to 3 significant figures.