Orthogonal projection w I am having trouble to solve this.

I know that $V=\{(x,y,z)\in \mathbb{R}^3 : x+y-z=0\}$ is a vector space in $\mathbb{R}^3$. Now consider a vector $u=(a,b,c)$ of $\mathbb{R}^3$ . Calculate the orthogonal projection $w$ of $u$ onto subspace $V$.

Note that I have found two basis vectors $v_1=(-1,1,0)$, and  $v_2= (1,0,1)$.
 A: Let
$$w = \lambda_1 v_1 + \lambda_2 v_2.\tag{1}$$
Note that $w \cdot v_1 = u\cdot v_1 = b-a$, and $w \cdot v_2 = u\cdot v_2 = a+c$. 
Using $(1)$, we get
$$w \cdot v_1 = b-a = 2\lambda_1 -\lambda_2, \tag{2}$$
and 
$$w \cdot v_2 = a+c = -\lambda_1 + 2 \lambda_2. \tag{3}$$
Solving $(2)$ and $(3)$ simultaneously results in
$$\lambda_1 = \frac{-a+2b+c}{3}, \tag{4}$$
and 
$$\lambda_2 = \frac{a+b+2c}{3}. \tag{5}$$
Therefore,
$$w = \lambda_1 v_1 + \lambda_2 v_2 = \left(\frac{2a-b+c}{3},\frac{-a+2b+c}{3},\frac{a+b+2c}{3}\right). \tag{6}$$
A: A natural  way of solving the question is as follows: We know that $(1,1,-1)$ is orthogonal to the plane. If we normalize this vector, we get 
$$
w=\left(\frac1{\sqrt3},\frac1{\sqrt3},-\frac1{\sqrt3}\right).
$$
The orthogonal projection onto the direction of $w$ is 
$$
P_wu=\langle u,w\rangle\,w=\frac{a+b-c}{\sqrt3}\,w=\left( \frac{a+b-c}3,\frac{a+b-c}3,\frac{-a-b+c}3\right).
$$
In matrix form,
$$
P_w=\frac13\,\begin{bmatrix}1&1&-1\\ 1&1&-1 \\ -1&-1&1 \end{bmatrix}.
$$
And the orthogonal projection onto $V$ is the orthogonal of $P_w$, that is $I-P_w$. So 
$$
P_V=I-P_w=\begin{bmatrix} 1&0&0\\0&1&0\\0&0&1\end{bmatrix} 
-\frac13\,\begin{bmatrix}1&1&-1\\ 1&1&-1 \\ -1&-1&1 \end{bmatrix}
=\frac13\,\begin{bmatrix}2&-1&1\\ -1&2&1 \\ 1&1&2 \end{bmatrix}
$$
Explicitly, 
$$
P_V(a,b,c)=\left(\frac{2a-b-c}3,\frac{-a+2b-c}3,\frac{a+b+2c}3 \right) 
$$
