Plane passing through $(2 \hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}}), \hat{\mathbf{i}}-\hat{\mathbf{j}}+\hat{\mathbf{k}}$ and $-\hat{\mathbf{i}}+\hat{\mathbf{j}}-\hat{\mathbf{k}}$. So, point in cartesian form are $(2,1,1),(1,-1,1)$ and $(-1,1,-1)$.
Equation of plane passing through three points
$$
\begin{aligned}
\left|\begin{array}{ccc}
x-x_1 & y-y_1 & z-z_1 \\
x_2-x_1 & y_2-y_1 & z_2-z_1 \\
x_3-x_1 & y_3-y_1 & z_3-z_1
\end{array}\right| & =0 \\
\left|\begin{array}{ccc}
x-2 & y-1 & z-1 \\
-1 & -2 & 0 \\
-3 & 0 & -2
\end{array}\right| & =0
\end{aligned}
$$
$$
\begin{array}{rlrl}
& \Rightarrow(x-2)(4)-(y-1)(2)+(z-1)(-6) & =0 \\
\Rightarrow & 4 x-8-2 y+2-6 z+6 & =0 \\
\Rightarrow & 4 x-2 y-6 z & =0 \\
\Rightarrow & 2 x-y-3 z & =0 \\
& \text { In vector form } & \mathbf{r} \cdot(2 \hat{\mathbf{i}}-\hat{\mathbf{j}}-3 \hat{\mathbf{k}}) & =0
\end{array}
$$

If $\mathbf{e}$ is unit vector perpendicular to given plane
So, $\quad \mathbf{e}=\frac{2 \hat{\mathbf{i}}-\hat{\mathbf{j}}-3 \hat{\mathbf{k}}}{\sqrt{2^2+(-1)^2+(-3)^2}}=\frac{1}{\sqrt{14}}(2 \hat{\mathbf{i}}-\hat{\mathbf{j}}-3 \hat{\mathbf{k}})$
Now, $\quad \mathbf{a}=2 \hat{\mathbf{i}}-3 \hat{\mathbf{j}}+6 \hat{\mathbf{k}}$

Projection of $\mathbf{a}$ on $\mathbf{e}=\mathbf{a} \cdot \mathbf{e}=\frac{4}{\sqrt{14}}+\frac{3}{\sqrt{14}}-\frac{18}{\sqrt{14}}$
$$
=\frac{-11}{\sqrt{14}} \quad\left[\because \mathbf{a} \cdot \mathbf{b}=a_1 a_2+b_1 b_2+c_1 c_2\right]
$$

Then, the projection vector of $\mathbf{a}$ on $\mathbf{e}$ is
$$
=-\frac{11}{\sqrt{14}}\left(\frac{1}{\sqrt{14}}(2 \hat{\mathbf{i}}-\hat{\mathbf{j}}-3 \hat{\mathbf{k}})\right)=\frac{11}{14}(-2 \hat{\mathbf{i}}+\hat{\mathbf{j}}+3 \hat{\mathbf{k}})
$$