### 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 skew lines are lines that are not parallel and do not intersect.
So I need to find conditions on $a$ that make these two lines neither parallel nor intersecting.
To check for parallelism, I can compare the direction vectors of the two lines.
If they are parallel, then one vector must be a scalar multiple of the other.
The direction vectors are $\begin{pmatrix} 2 \\ 3 \\ 4 \end{pmatrix}$ and $\begin{pmatrix} 5 \\ 2 \\ 1 \end{pmatrix}$.
If they are parallel, then there must be some scalar $k$ such that $\begin{pmatrix} 2 \\ 3 \\ 4 \end{pmatrix} = k \begin{pmatrix} 5 \\ 2 \\ 1 \end{pmatrix}$.
This means that $2 = 5k$, $3 = 2k$, and $4 = k$.
But these equations have no common solution for $k$, so the vectors are not parallel.
This means that the lines are either skew or intersecting, depending on the value of $a$.
To check for intersection, I can set the parametric equations of the two lines equal to each other and see if there is a common solution for $t$ and $u$.
The parametric equations are $x = 1 + 2t = 4 + 5u$, $y = 2 + 3t = 1 + 2u$, and $z = a + 4t = u$.
I can eliminate $u$ by using the equation for $z$.
I get $x = 1 + 2t = 4 + 5(a + 4t)$ and $y = 2 + 3t = 1 + 2(a + 4t)$.
I can simplify these equations by expanding and collecting like terms.
I get $x = 1 + 2t = 4 + 5a + 20t$ and $y = 2 + 3t = 1 + 2a + 8t$.
I can subtract the left sides from the right sides and get $0 = 3 + 5a + 18t$ and $0 = -1 + 2a + 5t$.
I can solve for $t$ by using the second equation.
I get $t = \frac{1 - 2a}{5}$.
I can plug this into the first equation and get $0 = 3 + 5a + 18(\frac{1 - 2a}{5})$.
I can simplify this equation by multiplying by 5 and expanding.
I get $0 = 15 + 25a + 18 - 36a$.
I can collect like terms and get $0 = 33 - 11a$.
I can solve for $a$ by dividing by -11.
I get $a = 3$.
This means that when $a = 3$, the two lines intersect at a common point.
Therefore, the lines are skew for any value of $a$ other than 3.