A line $L : \frac{x-2}{1}=\frac{y-3}{2}=\frac{z-1}{-1}$ is perpendicular to a plane (P), which passing through the point (4, 3, 9). If the mirror image of point 'S' on the line (L) in the given plane (P) is (2, 3, 1), then co-ordinates of point S, is :

(0, -1, 3)

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