A line L : (x-2)/1=(y-3)/2=(z-1)/(-1) is perpendicular to a plane (P)…

L : $\frac{x - 2}{1} = \frac{y - 3}{2} = \frac{z - 1}{-1}$ 

perpendicular to place

⇒  $\vec{n} = \hat{i} + 2\hat{j} - \hat{k}$

point in place = (4, 3, 9) ≡ $4\hat{i} + 3\hat{j} + 9\hat{k} = \vec{a}$

so  $\vec{r} - \vec{a}) . \vec{n} = 0$ → equation of place

so $\vec{r} . \vec{n} = \vec{a}) . \vec{n}$ 

$=(x\hat{i} + y\hat{j} + z\hat{k})(\hat{i}+2\hat{j}-\hat{k})$

$=(4\hat{i} + 3\hat{j} + 9\hat{k})(\hat{i}+2\hat{j}-\hat{k})$

= x + 2y - z 

= 4 + 6 - 9

⇒ x + 2y - z = 1

Let $\frac{x-2}{1} = \frac{y-3}{2} = \frac{z-1}{-1} = λ$

so x = λ + 2; y = 2λ + 3; z = -λ + 1 → point online in 'λ' form

Let S = S(λ +2, 2λ + 3, -λ + 1)

Q(2, 3, 1) are equidistant from place

so $\frac{|λ + 2 + 2(2λ + 3) + λ - 1 - 1|}{\sqrt{1^2+2^2+(-1)^2}} = \frac{|2+6-1-1|}{\sqrt{1^2+2^2 + (-1)^2}}$

= |λ + 2 + 4λ + 6 + λ - 1 - 1| = |6|

so |6λ + 6| = |6|

so |λ + 1| = |1|

so λ + 1 = ±1

so λ = 0, -2

at λ = 0

x = 2, y = 3, z = 1

at λ = -2

x = 0, y = -1, z = 3