Answer
\[
x+2 y-z=10
\]
Work Step by Step
Since the plane is perpendicular to the line
\[
-t=z, \quad 2 t-3=y, \quad t+2=x
\]
It's also perpendicular to the parallel vector of the line
\[
\langle 1,2,-1\rangle=\langle a, b, c\rangle=\vec{v}
\]
Since $\vec{v}$ is a normal plane vector and this plane passes through $p(-1,4,-3)$, the equation can be written as
\[
\begin{aligned}
0=& \left(x-x_{0}\right)a+\left(y-y_{0}\right)b+\left(z-z_{0}\right)c \\
\Rightarrow 0= &(x+1)+(y-4)2-(z+3) \\
\Rightarrow &0= 2 y-z+x-10
\end{aligned}
\]
We have found that the equation of the plane that satisfies the given stated conditions can be written as follows
\[
0=x+2 y-z-10
\]
\[
x+2 y-z=10
\]
