The existence of $\lambda$ has by now been proved by A.Kumar's answer
and the ensuing comments, but this still doesn't explain the behavior
of the determinant $p_\lambda(x,y,z)$: usually the zero-locus of
$3 \times 3$ determinant of linear forms is an irreducible curve,
not the union of three lines. A.Kumar's answer does strongly
hint at the reason, though. Here's an explanation, as well as
a formula for the number of $\lambda \bmod p$ for which
$p_\lambda$ has no nontrivial zeros.

Let $k$ be any field, $R$ the ring $k[T] \left/ (T^3 - \lambda T^2 - 1) \right.$,
and $r \in R$ the element $y+xT+zT^2$. Then multiplication by $r$
is a $k$-linear transformation of $R$ whose matrix with respect to
the basis $(T,1,T^2)$ is none other than
$$
\left(
\begin{array}{ccc}
x & y & z \\
y & z & x + \lambda z \\
z & x + \lambda z & y + \lambda x + \lambda ^2 z
\end{array}
\right)\phantom0.
$$
Thus $p_\lambda(x,y,z)$ is nonzero **iff** $r$ is invertible in $R$,
and this is true for all nonzero $r \in R$ **iff** $R$ is a field
**iff** the cubic $T^3 - \lambda T^2 - 1$ is irreducible.
Taking $T = 1/t$ then recovers A.Kumar's polynomial $t^3 + \lambda t + 1$.

Now as F.Brunault observes, if $k$ is finite then
the existence of a $\lambda$ that makes $t^3 + \lambda t + 1$
irreducible is an elementary counting argument: in degree $3$,
irreduicble still means no rational root; if $t^3 + \lambda t + 1 = 0$
then $t \neq 0$ and $\lambda = -(t^2 + t^{-1})$, so each nonzero $t$
arises exactly once as a root, and $t=0$ does not arise at all.
Since there are $\left|k\right|$ choices of $\lambda$, at least one
of them must make $t^3 + \lambda t + 1$ irreducible.

To enumerate such $\lambda$ we need only count the polynomials
$t^3 + \lambda t + 1$ that split completely. These are parametrized
by solutions of $t_1 + t_2 + t_3 = 0$, $t_1 t_2 t_3 = -1$ up to permutation.
These equations give an elliptic curve isogenous with the Fermat cubic
(if $a^3+b^3+c^3 = 0$ then
$$
(t_1^{\phantom.}, t_2^{\phantom.}, t_3^{\phantom.}) = \frac{-1}{abc} (a^3, b^3, c^3)
$$
satisfies $t_1 + t_2 + t_3 = 0$ and $t_1 t_2 t_3 = -1$; this gives
the isogeny in one direction). The number of rational points on
this curve is known, and eventually one finds (if I did this right)
that the count is $(p+1)/3$ if $p \equiv -1 \bmod 3$, and
$(p+1-a)/3$ if $p \equiv +1 \bmod 3$, where in the latter case
$a \equiv -1 \bmod 3$ is determined uniquely by $4p = a^2 + 27 b^2$.
In particular, for large $p$ the count is $p/3 + O(p^{1/2})$,
as predicted by Čebotarev (since $1/3$ is the proportion of
$3$-cycles in the symmetric group $S_3$).