I assume that you mean that point $Q$ has to be on the plane. There are many possible ways to approach this. Projections are probably the fastest way to do this, but let's just go full-algebraic on this one.
Let $Q = (q_1, q_2, q_3)$. Then we want to know how many scalar multiples of the normal vector we are away from $P$. That is, find $\lambda \in \mathbb{R}$ such that $Q + \lambda N = P$. This gives $(q_1 + \lambda, q_2 - 2\lambda, q_3 + \lambda) = (1,1,-1)$. This can be expanded into 3 linear equations. The $4$th equation comes from the condition that $Q$ should be in the plane: $q_1 - 2q_2 + q_3 = 4$. Solving all $4$ linear equations simultaneously should give you your point $Q$.
EDIT: I didn't address your method it seems. Doing $(Q - P) \times N = \vec{0}$ should work. Let's try it here:
\begin{align}
(q_1 - 1, q_2 - 1, q_3 + 1) \times (1,-2,1) &= (0,0,0)\\
(q_1 - 1, q_2 - 1, 2q_2 - q_1 + 5) \times (1,-2,1) &= (0,0,0) \text{ (sub plane equation)}\\
\left|\begin{array}{c c c} i & j & k \\ q_1 - 1 & q_2 - 1 & 2q_2 - q_1 + 5 \\
1 & -2 & 1 \end{array} \right| &= (0,0,0)\\
(5q_2 - 2q_1 + 9, 2q_2 - 2q_1 - 6, -2q_1 - q_2 + 3) &= (0,0,0)
\end{align}
Hence we have,
\begin{align}
5q_2 - 2q_1 + 9 &= 0\\
q_2 - q_1 + 3 &= 0\\
- q_2 - 2q_1 + 3 &= 0
\end{align}
Adding the second and third gives $-3q_1 + 6 = 0 \Rightarrow q_1 = 2$. Hence, $q_2 = 2 - 3 = -1$. For consistency, $5(-1) - 2(2) + 9 = -5 - 4 + 9 = 0$. Using the plane equation, we have which gives $q_3 = -(2) + 2(-1) + 4 = 0$. Finally, we have $Q = (2,-1,0)$. So surprise surprise, it looks like $\overline{PQ} = (1,-2,1)$ so that $\lambda = -1$ if you did it the first way I suggested.
EDIT 2: Thanks to amd, we can recognize that $Q = P = \lambda N$ for some $\lambda \in \mathbb{R}$, which we can put into the plane equation as:
\begin{align}
(P + \lambda N) \cdot N &= 4\\
P \cdot N + \lambda (N \cdot N) &= 4\\
(1,1,-1) \cdot (1,-2,1) + \lambda (1,-2,1) \cdot (1,-2,1) &= 4\\
-2 + 6 \lambda &= 4\\
\lambda &= 1
\end{align}
Hence, $Q = (1,1,-1) + (1,-2,1) = (2,-1,0)$.