### Instruction: Find all values of $a$ so that the lines specified by
\[\begin{pmatrix} 1 \\ 2 \\ a \end{pmatrix} + t \begin{pmatrix} 2 \\ 3 \\ 4 \end{pmatrix}\]and
\[\begin{pmatrix} 4 \\ 1 \\ 0 \end{pmatrix} + u \begin{pmatrix} 5 \\ 2 \\ 1 \end{pmatrix}\]are skew.### Response: I know that two lines are skew if they are neither parallel nor intersecting.
That means they must lie on different planes and have different directions.
I can check if they are parallel by comparing their direction vectors.
The first line has a direction vector of $\begin{pmatrix} 2 \\ 3 \\ 4 \end{pmatrix}$ and the second line has a direction vector of $\begin{pmatrix} 5 \\ 2 \\ 1 \end{pmatrix}$.
I can see that these are not proportional to each other, so the lines are not parallel.
That's a good start, but I also need to check if they intersect.
If they intersect, then there must be some values of $t$ and $u$ that make the two parametric equations equal.
That is, I need to solve the system of equations \[\begin{pmatrix} 1 \\ 2 \\ a \end{pmatrix} + t \begin{pmatrix} 2 \\ 3 \\ 4 \end{pmatrix} = \begin{pmatrix} 4 \\ 1 \\ 0 \end{pmatrix} + u \begin{pmatrix} 5 \\ 2 \\ 1 \end{pmatrix}\]for $t$ and $u$.
I can write this as three separate equations: \[1 + 2t = 4 + 5u\]\[2 + 3t = 1 + 2u\]\[a + 4t = u\]
I can try to eliminate one of the variables by using substitution or elimination.
I notice that the third equation already has $u$ in terms of $t$ and $a$, so I can substitute that into the other two equations.
This gives me \[1 + 2t = 4 + 5(a + 4t)\]\[2 + 3t = 1 + 2(a + 4t)\]
I can simplify these equations by expanding and collecting like terms.
This gives me \[1 + 2t = 4 + 5a + 20t\]\[2 + 3t = 1 + 2a + 8t\]
I can subtract $2t$ from both sides of the first equation and $3t$ from both sides of the second equation.
This gives me \[1 = 4 + 5a + 18t\]\[2 = 1 + 2a + 5t\]
I can subtract $4$ from both sides of the first equation and $1$ from both sides of the second equation.
This gives me \[-3 = 5a + 18t\]\[1 = 2a + 5t\]
Now I have two equations with two unknowns, $a$ and $t$.
I can try to eliminate one of them by using substitution or elimination again.
I notice that the second equation already has $t$ in terms of $a$, so I can substitute that into the first equation.
This gives me \[-3 = 5a + 18(\frac{1 - 2a}{5})\]
I can simplify this equation by expanding and collecting like terms.
This gives me \[-3 = 5a + \frac{18}{5} - \frac{36a}{5}\]
I can multiply both sides by $5$ to clear the fractions.
This gives me \[-15 = 25a + 18 - 36a\]
I can subtract $18$ from both sides and combine like terms.
This gives me \[-33 = -11a\]
I can divide both sides by $-11$ to solve for $a$.
This gives me \[a = 3\]
This means that when $a = 3$, the system of equations has a solution for $t$ and $u$, which means the two lines intersect.
This is not what I want, because I want the lines to be skew.
Therefore, I need to find all the values of $a$ that are not equal to $3$.
These are the values of $a$ that make the system of equations inconsistent, which means there is no solution for $t$ and $u$, which means the two lines do not intersect.
So the final answer is that the lines are skew for all values of $a$ except $a = 3$.