Equation of the plane $\pi_1$ passing through the three non-collinear points $(0,1,4,(1,0,-4)$ and $(-2,1,0)$ is
$$
\begin{array}{rlc}
& & \left|\begin{array}{ccc}
x-0 & y-1 & z-2 \\
1-0 & 0-1 & -2-2 \\
-2-0 & 1-1 & 0-2
\end{array}\right|=0 \\
\Rightarrow & \left|\begin{array}{ccc}
x & y-1 & z-2 \\
1 & -1 & -4 \\
-2 & 0 & -2
\end{array}\right|=0 \\
\Rightarrow & x(2-0)-(y-1)(-2-8)+(z-2)(0-2)=0 \\
\Rightarrow & 2 x-(y-1) \times(-10)-2(z-2)=0 \\
\Rightarrow & 2 x+10 y-10-2 z+4=0 \\
\Rightarrow & 2 x+10 y-2 z-6=0
\end{array}
$$
Now, equation of the plane $\pi_2$ passing through $(1,2,3)$, is
$$
a(x-1)+b(y-4)+c(z-3)=0
$$
$\because$ Plane (i) is perpendicular to $x+y+z=1$
$$
\therefore \quad a+b+c=0
$$
Also, plane (i) is perpendicular to the plane
$$
\begin{aligned}
& 2 x-3 y+z=5 \\
& \therefore \quad 2 a-3 b+c=0
\end{aligned}
$$
From Eqs. (ii) and (iii), we get
$$
\begin{aligned}
& \frac{a}{1+3}=\frac{b}{2-1}=\frac{c}{-3-2}=\lambda \text { (Iet) } \\
& \Rightarrow \quad a=4 \lambda, b=\lambda \text { and } c=-5 \lambda \\
&
\end{aligned}
$$
On substituting these values in Eq. (i), we get


$$
\begin{array}{rlrl}
& 4 \lambda(x-1)+\lambda(y-2)-5 \lambda(z-3) & =0 \\
\Rightarrow & & 4 x-4+y-2-5 z+15 & =0 \\
\Rightarrow & & 4 x+y-5 z+9 & =0
\end{array}
$$
which is the required equation of the plane $\pi_2$. Now, acute angle between the planes $\pi_1$ and $\pi_2$, is given as
$$
\begin{aligned}
\cos \theta & =\left|\frac{2 \times 4+10 \times 1+(-4) \times(-5)}{\sqrt{2^2+10^2+\left(-2^2\right.} \sqrt{4^2+1^2+(-5)^2}}\right| \\
& =\left|\frac{8+10+10}{\sqrt{108} \sqrt{16+1+25}}\right| \\
& =\left|\frac{28}{\sqrt{2 \times 2 \times 3 \times 3 \times 3} \sqrt{2 \times 3 \times 7}}\right| \\
& =\left|\frac{28}{2 \times 3 \times 3 \sqrt{14}}\right|=\frac{14}{9 \sqrt{14}}=\frac{\sqrt{14}}{9}
\end{aligned}
$$