To find the vector perpendicular to the plane containing the vectors $$\hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}}$$ and $$\hat{\mathbf{i}}+2\hat{\mathbf{j}}+3\hat{\mathbf{k}}$$, we can use the cross product of these two vectors.

The cross product is defined as:

$$\mathbf{A} \times \mathbf{B} = (a_2b_3 - a_3b_2)\hat{\mathbf{i}} + (a_3b_1 - a_1b_3)\hat{\mathbf{j}} + (a_1b_2 - a_2b_1)\hat{\mathbf{k}}$$

For our vectors, this becomes:

$$\begin{aligned} (\hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}}) \times (\hat{\mathbf{i}}+2\hat{\mathbf{j}}+3\hat{\mathbf{k}}) &= \left|\begin{array}{ccc} \hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ 1 & 1 & 1 \\ 1 & 2 & 3 \\ \end{array}\right| \\ &= \hat{\mathbf{i}}(1 \cdot 3 - 1 \cdot 2) - \hat{\mathbf{j}}(1 \cdot 3 - 1 \cdot 1) + \hat{\mathbf{k}}(1 \cdot 2 - 1 \cdot 1) \\\\ &= \hat{\mathbf{i}}(3 - 2) - \hat{\mathbf{j}}(3 - 1) + \hat{\mathbf{k}}(2 - 1) \\\\ &= \hat{\mathbf{i}}(1) - \hat{\mathbf{j}}(2) + \hat{\mathbf{k}}(1) \\\\ &= \hat{\mathbf{i}} - 2\hat{\mathbf{j}} + \hat{\mathbf{k}}. \end{aligned}$$

Now that we have the perpendicular vector, let us compute the magnitude of the projection of the vector $$2\hat{\mathbf{i}}+ 3\hat{\mathbf{j}}+ \hat{\mathbf{k}}$$ onto $$\hat{\mathbf{i}} - 2\hat{\mathbf{j}} + \hat{\mathbf{k}}$$.

The projection of vector $$\mathbf{a}$$ onto vector $$\mathbf{b}$$ is given by the formula:

$$ \text{proj}_{\mathbf{b}}(\mathbf{a}) = \frac{\mathbf{a} \cdot \mathbf{b}}{\|\mathbf{b}\|} $$

But we are only interested in the magnitude of the projection, which simplifies to:

$$ \|\text{proj}_{\mathbf{b}}(\mathbf{a})\| = \left| \frac{\mathbf{a} \cdot \mathbf{b}}{\|\mathbf{b}\|} \right| $$

Let's find the dot product of vectors $$2\hat{\mathbf{i}}+ 3\hat{\mathbf{j}}+ \hat{\mathbf{k}}$$ and $$\hat{\mathbf{i}} - 2\hat{\mathbf{j}} + \hat{\mathbf{k}}$$:

$$\begin{aligned} \mathbf{a} \cdot \mathbf{b} &= (2\hat{\mathbf{i}}+ 3\hat{\mathbf{j}}+ \hat{\mathbf{k}}) \cdot (\hat{\mathbf{i}} - 2\hat{\mathbf{j}} + \hat{\mathbf{k}}) \\\\ &= 2(1) - 3(2) + 1(1) \\\\ &= 2 - 6 + 1 \\\\ &= -3. \end{aligned}$$

Next, let's find the magnitude of $$\mathbf{b}$$, which is $$\|\mathbf{b}\|$$:

$$\begin{aligned} \|\mathbf{b}\| &= \sqrt{(\hat{\mathbf{i}} - 2\hat{\mathbf{j}} + \hat{\mathbf{k}}) \cdot (\hat{\mathbf{i}} - 2\hat{\mathbf{j}} + \hat{\mathbf{k}})} \\\\ &= \sqrt{1^2 + (-2)^2 + 1^2} \\\\ &= \sqrt{1 + 4 + 1} \\\\ &= \sqrt{6}. \end{aligned}$$

Finally, the magnitude of the projection is:

$$\begin{aligned} \|\text{proj}_{\mathbf{b}}(\mathbf{a})\| &= \left| \frac{\mathbf{a} \cdot \mathbf{b}}{\|\mathbf{b}\|} \right| \\\\ &= \left| \frac{-3}{\sqrt{6}} \right| \\\\ &= \left| -\frac{\sqrt{9}}{\sqrt{6}} \right| \\\\ &= \left| -\sqrt{\frac{3}{2}} \right| \\\\ &= \sqrt{\frac{3}{2}} \end{aligned}$$

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