Find the value of the constant a and the position vector of the point of intersection of two lines.

Given the equations of the lines:

Equating the $\mathbf{i}$, $\mathbf{j}$, and $\mathbf{k}$ components:

1. $1+\lambda = a$
2. $a+2\lambda = -\mu$
3. $-1+2\lambda = 2+3\mu$

Substitute equation (1) into equation (2):

From equation (4), we can express $\mu$ in terms of $\lambda$:

Substitute this expression for $\mu$ into equation (3):

Add $9\lambda$ to both sides:

Now substitute $\lambda=0$ back into equation (1) to find $a$:

The value of the constant $a$ is $\boxed{1}$.

5. (b) The position vector of the point of intersection

Using $\lambda=0$ and $a=1$ in the equation for $\mathbf{r_1}$:

The position vector of the point of intersection is $\boxed{\mathbf{i} + \mathbf{j} - \mathbf{k}}$.

5. (c) The cosine of the acute angle between $\mathbf{r_1}$ and $\mathbf{r_2}$

The direction vector for $\mathbf{r_1}$ is $\mathbf{d_1} = \mathbf{i} + 2\mathbf{j} + 2\mathbf{k}$.
The direction vector for $\mathbf{r_2}$ is $\mathbf{d_2} = -\mathbf{j} + 3\mathbf{k}$.

The cosine of the angle $\theta$ between two vectors is given by $\cos \theta = \frac{\mathbf{d_1} \cdot \mathbf{d_2}}{|\mathbf{d_1}||\mathbf{d_2}|}$. For the acute angle, we take the absolute value of the dot product.

Rationalize the denominator:

The cosine of the acute angle between $\mathbf{r_1}$ and $\mathbf{r_2}$ is $\boxed{\frac{2\sqrt{10}}{15}}$.