QJEEM21Ma0841: JEE Main 2021 Mathematics - Class 12 JEEM21Ma0841 Solution: The distance of line from the point (2, 1, 6) is :

$3y - 2z - 1 = 0 = 3x - z + 4$

$3y - 2z - 1 = 0$

D.R's $\Rightarrow$ (0, 3, $-$ 2)

$3x - z + 4$ = 0

D.R's $\Rightarrow$ (3, $-$ 1, 0)

Let DR's of given line are a, b, c

Now, 3b $-$ 2c = 0 & 3a $-$ c = 0

$\therefore$ 6a = 3b = 2c

a : b : c = 3 : 6 : 9

Any point on line

3K $-$ 1, 6K + 1, 9K + 1

Now, 3(3K $-$ 1) + 6(6K + 1)1 + 9(9K + 1) = 0

$\Rightarrow$ K = ${1 \over 3}$

Point on line $\Rightarrow$ (0, 3, 4)

Given point (2, $-$ 1, 6)

$\Rightarrow$ Distance = $\sqrt {4 + 16 + 4} = 2\sqrt 6$

Option (c)