Given points are $A(-2,1,3), B(1,1,1)$ and $C(2,3,4)$.
$ \therefore A =-2 \hat{ i }+\hat{ j }+3 \hat{ k }, B =\hat{ i }+\hat{ j }+\hat{ k } $ and
$C =2 \hat{ i }+3 \hat{ j }+4 \hat{ k }$
Now, $ A B =(\hat{ i }+\hat{ j }+\hat{ k })-(-2 \hat{ i }+\hat{ j }+\hat{ k })=3 \hat{ i }-2 \hat{ k } $
$ B C =(2 \hat{ i }+3 \hat{ j }+4 \hat{ k })-(\hat{ i }+\hat{ j }+\hat{ k })=\hat{ i }+2 \hat{ j }+3 \hat{ k }$
The normal of the plane $n = A B \times B C$
$\therefore A B \times B C =\begin{vmatrix}\hat{ i } & \hat{ j } & \hat{ k } \\ 3 & 0 & -2 \\ 1 & 2 & 3\end{vmatrix}$
$ = \hat{i} (0 + 4) - \hat{j} (9 + 2) + \hat{k}(6 - 0)$
$n = 4\hat{i} - 11\hat{j} + 6\hat{k}$
Now, equation of plane
$\Rightarrow 4(x-2)-11(y-3)+6(z-4)=0$
$\Rightarrow 4 x-8-11 y+33+6 z-24=0 $
$\Rightarrow 4 x-11 y+6 z+1=0$
As we know that, if $a x+b y+c z+d=0$
is the equation of plane, then the normal form is
$\left(\frac{a}{\sqrt{a^{2}+b^{2}+c^{2}}}\right) x+\left(\frac{b}{\sqrt{a^{2}+b^{2}+c^{2}}}\right) y$
$+\left(\frac{c}{\sqrt{a^{2}+b^{2}+c^{2}}}\right) z+\left(\frac{d}{\sqrt{a^{2}+b^{2}+c^{2}}}\right)=0$
$\therefore $ Required normal form is
$\left(\frac{4}{\sqrt{4^{2}+11^{2}+6^{2}}}\right) x+\left(\frac{-11}{\sqrt{4^{2}+11^{2}+6^{2}}}\right) y$
$+\left(\frac{6}{\sqrt{4^{2}+11^{2}+6^{2}}}\right) z+\left(\frac{1}{\sqrt{4^{2}+11^{2}+6^{2}}}\right)=0$
$\Rightarrow \frac{4 x}{\sqrt{16+12 l+36}}-\frac{11 y}{\sqrt{16+12 l+36}}$
$+\frac{6 z}{\sqrt{16+121+36}}+\frac{1}{\sqrt{16+121+36}}=0$
$\Rightarrow \frac{4 x}{\sqrt{173}}-\frac{11 y}{\sqrt{173}}+\frac{6 z}{\sqrt{173}}+\frac{1}{\sqrt{173}}=0$
$\Rightarrow \frac{4 x}{\sqrt{173}}-\frac{11 y}{\sqrt{173}}+\frac{6 z}{\sqrt{173}}=-\frac{1}{\sqrt{173}}$
$\Rightarrow -\frac{4 x}{\sqrt{173}}+\frac{11 y}{\sqrt{173}}-\frac{6 z}{\sqrt{173}}=\frac{1}{\sqrt{173}}$
$\Rightarrow \left(-\frac{4}{\sqrt{173}}\right) x+\left(\frac{11}{\sqrt{173}}\right) y+\left(\frac{-6}{\sqrt{173}}\right) z=\frac{1}{\sqrt{173}}$