You're computing $u_2$ wrongly.
I find it useful to set up a systematic way, where the information can be picked up easily.
Let $v_1$, $v_2$ and $v_3$ be the three columns of $A$.
GS1
$u_1=v_1$
$\langle u_1,u_1\rangle=2$
GS2
$\alpha_{12}=\dfrac{\langle u_1,v_2\rangle}{\langle u_1,u_1\rangle}=\dfrac{1}{2}$
$u_2=v_2-\alpha_{12}u_1=\begin{bmatrix}1/2\\1\\-1/2\end{bmatrix}$
$\langle u_2,u_2\rangle=\dfrac{3}{2}$
GS3
$\alpha_{13}=\dfrac{\langle u_1,v_3\rangle}{\langle u_1,u_1\rangle}=\dfrac{1}{2}$
$\alpha_{23}=\dfrac{\langle u_2,v_3\rangle}{\langle u_2,u_2\rangle}=\dfrac{1}{3}$
$u_3=v_3-\alpha_{13}u_1-\alpha_{23}u_2=\begin{bmatrix}-2/3\\2/3\\2/3\end{bmatrix}$
$\langle u_3,u_3\rangle=\dfrac{4}{3}$
Matrix $Q$
The matrix $Q$ has as columns the vectors $u_1$, $u_2$ and $u_3$ divided by their norms:
$$
Q=\begin{bmatrix}
1/\sqrt{2} & 1/\sqrt{6} & -1/\sqrt{3} \\
0 & 2/\sqrt{6} & 1/\sqrt{3} \\
1/\sqrt{2} & -1/\sqrt{6} & 1/\sqrt{3}
\end{bmatrix}
$$
Matrix $R$
The matrix $R$ is obtained by multiplying each row of the upper unitriangular with the entries $\alpha_{ij}$ by the norm of the corresponding $u$ vector.
$$
R=\begin{bmatrix}
1 & 1/2 & 1/2 \\
0 & 1 & 1/3 \\
0 & 0 & 1
\end{bmatrix}
\begin{array}{l}\cdot\sqrt{2}\\\cdot \sqrt{6}/2\\\cdot 2/\sqrt{3}\end{array}=
\begin{bmatrix}
\sqrt{2} & \sqrt{2}/2 & \sqrt{2}/2 \\
0 & \sqrt{6}/2 & \sqrt{6}/6 \\
0 & 0 & 2/\sqrt{3}
\end{bmatrix}
$$
Since the notation is sometimes different: In the following I assume $T_{B',B}$ takes a vector in its representation w.r.t. $B$ and evaluates $T$ and gives back a vector in representation w.r.t. $B'$. (Sometimes the order of $B$ and $B'$ is different.)
The idea in (c) is to change basis from $B$ to $B'$ and evaluate T after that. Try to calculate the matrix $T_{B',B'}$ first. After that you can calculate the base change matrix $M_{B',B}$, changing the representation of $x$ w.r.t. $B$ into the representation of $x$ w.r.t. $B'$. Then the answer is the product of these matrices (Which one do you have to multiply with $x$ first? So which order is the right one?).
For (d) you just need to evaluate T on (1,1,1), i. e. multiply the matrix in (c) from right with (1,1,1).
Best Answer
Simple test of row reduction will show that subspace V has dimension 2 which is a plane. $v_1,v_2$ can be taken a basis, take a cross-product $v_1\times v_2 = (1,-1,1)$ which is the normal('s direction ratios). Equation of plane that represents $V$ is given by $x-y+z=0$ and using distance formula $$d=\left|\frac{ax_1+by_1+cz_1-d}{\sqrt{a^2+b^2+c^2}} \right| = \frac{1-2+3}{\sqrt{3}}=\frac{2}{\sqrt3}$$