Find the equation of plane passing through $A(3,-2,-1)$ and parallel to the vectors $\overline{b}=\overline{i}-2\overline{j}+4\overline{k},\overline{c}=3\overline{i}+2\overline{j}-5\overline{k}.$

Given that $A(3,-2,-1)$

$\therefore \overline{a}=\overline{OA}=3\overline{i}-2\overline{j}-\overline{k}$

$\overline{b}=\overline{i}-2\overline{j}+4\overline{k};\overline{c}=3\overline{i}+2\overline{j}-5\overline{k}$

$\therefore$ The vector equation of the plane passing through the points A and parallel to the vectors $\overline{b},\overline{c}$ is

$\left[\begin{array}{l}\overline{r}-\overline{a}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\overline{b}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\overline{c}\end{array}\right]=0$

$\Rightarrow [\overline{r}-(3\overline{i}-2\overline{j}-\overline{k}) \overline{i}-2\overline{j}+4\overline{k} 3\overline{i}+2\overline{j}-5\overline{k}]=0$

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

The Cartesian eq. of the plane passing through A and parallel to $\overline{b}$ and $\overline{c}$ is

$\left|\begin{array}{ccc}x-3& y+2& z+1\\ 1& -2& 4\\ 3& 2& -5\end{array}\right|=0$

$\Rightarrow (x-3)(10-8)-(y+2)(-5-12)+(z+1)(2+6)=0$

$\Rightarrow (x-3)2-(y+2)(-17)+(z+1)8=0$

$\Rightarrow 2x-6+17y+34+8z+8=0$

$\Rightarrow 2x+17y+8z+36=0$