The distance of line

Solution:

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

$3 \mathrm{y}-2 \mathrm{z}-1=0 \quad$ D.R's $\Rightarrow(0,3,-2)$

$3 x-z+4=0 \quad$ D.R's $\Rightarrow(3,-1,0)$

Let DR's of given line are $\mathrm{a}, \mathrm{b}, \mathrm{c}$

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

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

$a: b: c=3: 6: 9$

Any pt on line

$3 \mathrm{~K}-1,6 \mathrm{~K}+1,9 \mathrm{~K}+1$

Now $3(3 \mathrm{~K}-1)+6(6 \mathrm{~K}+1) 1+9(9 \mathrm{~K}+1)=0$

$\Rightarrow \mathrm{K}=\frac{1}{3}$

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

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

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

Option (3)