Find the equation of the plane passing through points $A(2, 3, –1), B(4, 5, 2), C(3, 6, 5).$

Given points are $A(2,3,-1),B(4,5,2),C(3,6,5)$

$\therefore \overline{a}=\overline{OA}=2\overline{i}+3\overline{j}-\overline{k}, \overline{b} =\overline{OB}=4\overline{i}+5\overline{j}+2\overline{k}$

$\overline{c}=\overline{OC}=3\overline{i}+6\overline{j}+5\overline{k}$

$\therefore \overline{b}-\overline{a}=(4\overline{i}+5\overline{j}+2\overline{k})-(2\overline{i}+3\overline{j}-\overline{k})=2\overline{i}+2\overline{j}+3\overline{k}.$

$\overline{c}-\overline{a}=\left(3\hat{i}+6\hat{j}+5\hat{k}\right)-\left(2\hat{i}+3\hat{j}-\hat{k}\right)=\hat{i}+3\hat{j}+6\hat{k}$

The vector eq. of the plane passing through the given points is $\left[\begin{array}{ccc}\overline{r}-\overline{a}& \overline{b}-\overline{a}& \overline{c}-\overline{a}\end{array}\right] =0$

$\Rightarrow \left[\begin{array}{ccc}\overline{r}-(2\overline{i}+3\overline{j}-\overline{k})& 2\overline{i}+2\overline{j}+3\overline{k}& \overline{i}+3\overline{j}+6\overline{k}\end{array}\right]=0$

Now $A=\left(x_1,y_1,z_1\right)=(2,3,-1)$

$B=\left(x_2,y_2,z_2\right)=(4,5,2)$

$C=\left(x_3,y_3,z_3\right)=(3,6,5)$

$\therefore$ The Cartesian eq. of the plane passing through given points is $\left|\begin{array}{ccc}x-x_1& y-y_1& z-z_1\\ x_2-x_1& y_2-y_1& z_2-z_1\\ x_3-x_1& y_3-y_1& z_3-z_1\end{array}\right|=0$

$\Rightarrow \left|\begin{array}{l}x-2\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}y-3\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}z+1\\ 4-2\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}5-3\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}2+1\\ 3-2\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}6-3\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}5+1\end{array}\right|=0$

$\Rightarrow \left|\begin{array}{ccc}x-2& y-3& z+1\\ 2& 2& 3\\ 1& 3& 6\end{array}\right|=0$

$\Rightarrow (x-2)\left(3\right)-(y-3)9+(z+1)4=0$

$3x-9y+4z+25=0$