Find the vector and cartesian equations of the planes

(a) that passes through the point $(1,0, - 2)$ and the normal to the plane is $(\hat i + \hat j - \hat k)$ .
(b) that passes through the point $(1,4,6)$and the normal vector to the plane is $\hat i - 2\hat j + \hat k$

. : (a) Vector Form :
The equation of the plane,

which passes through $\vec a$ and normal to$\vec n$ is $(\vec r - \vec a) \cdot \vec n = 0$

Here, $\vec a(1,0, - 2) = \hat i - 2\hat k$ and $\vec n = \hat i + \hat j - \hat k$

therefore the equation of the plane is,
$[\vec r - (\hat i - 2\hat k)] \cdot (\hat i + \hat j - \hat k) = 0 \Rightarrow \vec r \cdot (\hat i + \hat j - \hat k) = 3$

Cartesian Form :
If $< a,b,c >$ are direction ratios of the normal to the plane,

then the equation of the plane through $({x_1},{y_1},{z_1})$ is
$a(x - {x_1}) + b(y - {y_1}) + c(z - {z_1}) = 0.$

This passes through $(1,0, - 2).$
$\therefore$ ${x_1} = 1,{y_1} = 0,{z_1} = - 2$

Direction ratios of normal $\hat i + \hat j - \hat k$ are $< 1,1, - 1 >$
$\therefore$ $a = 1,b = 1,c = - 1$

The equation of the plane is,
$1 \cdot (x - 1) + 1 \cdot (y - 0) - 1(z + 2) = 0$

$\Rightarrow$ $x - 1 + y - z - 2 = 0 \Rightarrow x + y - z = 3$

(b) Vector Form :
Here, $\vec a(1,4,6) = \hat i + 4\hat j + 6\hat k$ and $\vec n = \hat i - 2\hat j + \hat k$

$\therefore$ The equation of the plane is :
$(\vec r - \vec a) \cdot \vec n = 0 \Rightarrow [\vec r - (\hat i + 4\hat j + 6\hat k)] \cdot (\hat i - 2\hat j + \hat k) = 0$

$\Rightarrow$ $\vec r \cdot (\hat i - 2\hat j + \hat k) + 1 = 0$

Cartesian Form :

If $(a,b,c)$ are direction ratios of the normal to the plane,

then the equation of the plane through $({x_1},{y_1},{z_1})$ is
$a(x - {x_1}) + b(y - {y_1}) + c(z - {z_1}) = 0$

This passes through $(1,4,6)$
$\therefore$ ${x_1} = 1,{y_1} = 4,{z_1} = 6$

Direction-ratios of normal $\hat i - 2\hat j + \hat k$ are $< 1, - 2,1 >$

$\therefore$ $a = 1,b = - 2,c = 1$

$\therefore$ The equation of the plane is
$1(x - 1) - 2(y - 4) + 1(z - 6) = 0$

$\Rightarrow$ $x - 2y + z - 1 + 8 - 6 = 0$ $\Rightarrow$ $x - 2y + z + 1 = 0$