If the straight lines $\dfrac{x-2}{1}=\dfrac{y-3}{1}=\dfrac{z-4}{-t}$ and $\dfrac{x-1}{t}=\dfrac{y-4}{2}=\dfrac{z-5}{1}$ are intersecting then $t$ can have

The given lines are $L_1: \dfrac{x-2}{1} = \dfrac{y-3}{1} = \dfrac{z-4}{-t} = \lambda$ and $L_2: \dfrac{x-1}{t} = \dfrac{y-4}{2} = \dfrac{z-5}{1} = \mu$.

The general points on the lines are $P_1 = (2+\lambda, 3+\lambda, 4-t\lambda)$ and $P_2 = (1+t\mu, 4+2\mu, 5+\mu)$.

For the lines to intersect, there must exist $\lambda$ and $\mu$ such that $P_1 = P_2$. This gives the system of equations:

1) $2+\lambda = 1+t\mu \Rightarrow \lambda - t\mu = -1$

2) $3+\lambda = 4+2\mu \Rightarrow \lambda - 2\mu = 1$

3) $4-t\lambda = 5+\mu \Rightarrow t\lambda + \mu = -1$

Subtracting equation (2) from (1): $(-t+2)\mu = -2 \Rightarrow \mu = \dfrac{-2}{2-t} = \dfrac{2}{t-2}$.

Substituting $\mu$ into (2): $\lambda = 1 + 2\mu = 1 + \dfrac{4}{t-2} = \dfrac{t+2}{t-2}$.

Substituting $\lambda$ and $\mu$ into (3): $t\left(\dfrac{t+2}{t-2}\right) + \dfrac{2}{t-2} = -1$.

$t^2 + 2t + 2 = -(t-2) = -t + 2$.

$t^2 + 3t = 0 \Rightarrow t(t+3) = 0$.

The possible values for $t$ are $0$ and $-3$. Thus, there are exactly two values.

Answer: Exactly two values