Solve in english

If two lines are perpendicular, their direction vectors are orthogonal, meaning their dot product is zero: $\mathbf{d_1} \cdot \mathbf{d_2} = 0$.

$$(2)(3) + (-1)(-2) + (a)(-2) = 0$$ $$6 + 2 - 2a = 0$$ $$8 - 2a = 0$$

$$2a = 8$$ $$a = \boxed{4}$$

Substitute $a=4$ into the equation for $L_1$:

$$\mathbf{i} - 2\mathbf{j} + 3\mathbf{k} + \lambda(2\mathbf{i} - \mathbf{j} + 4\mathbf{k}) = -\mathbf{i} - \mathbf{j} - \mathbf{k} + \mu(3\mathbf{i} - 2\mathbf{j} - 2\mathbf{k})$$

Equate the components:
For $\mathbf{i}$: $1 + 2\lambda = -1 + 3\mu \implies 2\lambda - 3\mu = -2 \quad (1)$
For $\mathbf{j}$: $-2 - \lambda = -1 - 2\mu \implies -\lambda + 2\mu = 1 \quad (2)$
For $\mathbf{k}$: $3 + 4\lambda = -1 - 2\mu \implies 4\lambda + 2\mu = -4 \implies 2\lambda + \mu = -2 \quad (3)$

From equation (2), we can express $\lambda$ in terms of $\mu$:

$$\lambda = 2\mu - 1$$

Substitute this into equation (3):

$$2(2\mu - 1) + \mu = -2$$ $$4\mu - 2 + \mu = -2$$ $$5\mu = 0$$ $$\mu = 0$$

Now substitute $\mu=0$ back into the expression for $\lambda$:

$$\lambda = 2(0) - 1$$ $$\lambda = -1$$

Verify these values with equation (1): $2(-1) - 3(0) = -2 - 0 = -2$. The values are consistent.

Using $L_1$ with $\lambda = -1$:

$$\mathbf{r} = \mathbf{i} - 2\mathbf{j} + 3\mathbf{k} + (-1)(2\mathbf{i} - \mathbf{j} + 4\mathbf{k})$$ $$\mathbf{r} = \mathbf{i} - 2\mathbf{j} + 3\mathbf{k} - 2\mathbf{i} + \mathbf{j} - 4\mathbf{k}$$ $$\mathbf{r} = (1-2)\mathbf{i} + (-2+1)\mathbf{j} + (3-4)\mathbf{k}$$ $$\mathbf{r} = \boxed{-\mathbf{i} - \mathbf{j} - \mathbf{k}}$$

The plane contains both lines, so its normal vector $\mathbf{n}$ must be perpendicular to both direction vectors $\mathbf{d_1}$ and $\mathbf{d_2}$. We can find $\mathbf{n}$ by taking the cross product of $\mathbf{d_1}$ and $\mathbf{d_2}$.
Using $a=4$, $\mathbf{d_1} = \begin{pmatrix} 2 \\ -1 \\ 4 \end{pmatrix}$ and $\mathbf{d_2} = \begin{pmatrix} 3 \\ -2 \\ -2 \end{pmatrix}$.

$$\mathbf{n} = \mathbf{d_1} \times \mathbf{d_2} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 2 & -1 & 4 \\ 3 & -2 & -2 \end{vmatrix}$$ $$\mathbf{n} = \mathbf{i}((-1)(-2) - (4)(-2)) - \mathbf{j}((2)(-2) - (4)(3)) + \mathbf{k}((2)(-2) - (-1)(3))$$ $$\mathbf{n} = \mathbf{i}(2 - (-8)) - \mathbf{j}(-4 - 12) + \mathbf{k}(-4 - (-3))$$ $$\mathbf{n} = 10\mathbf{i} + 16\mathbf{j} - \mathbf{k}$$

The Cartesian equation of a plane is $Ax + By + Cz = D$, where $\begin{pmatrix} A \\ B \\ C \end{pmatrix}$ is the normal vector.
So, the equation is $10x + 16y - z = D$.
We know the point of intersection $P(-1, -1, -1)$ lies on the plane. Substitute these coordinates to find $D$:

$$10(-1) + 16(-1) - (-1) = D$$ $$-10 - 16 + 1 = D$$ $$-25 = D$$