This is a plane, not a line!

The equation can be rewritten as $((x,y,z)-(1,2,3))\cdot(1,\dfrac 12,\dfrac 13)=0$
which means that the plane passes through $(1,2,3)$ and $(1,\dfrac 12,\dfrac 13)$ is a normal vector.
$6(1,\dfrac 12,\dfrac 13)=(6,3,2)$ is also a normal vector.

Multiplying through by $6$ we get: $6x-6+3y-6+2z-6=0 \Rightarrow 6x+3y+2x=18$

A normal vector to the plane $ax+by+cz=d$ is $(a,b,c)$ so the answer is:

$\boxed{(6,3,2)}$

This comes to 6(x-1) + 3(y-2) + 2(z-3) = 0. Hence, the normal vector will be (6,3,2), meeting the plane at point (1,2,3)

$\frac{6(x-1) + 3(y-2) + 2(z-3)}{6}$ =0 so coefficient of x is 6, of y is 3 and of z is 2. so the normal vector will be (6,3,2).

LCM is 6, Denom of 1 means 6, 2 means 3 and 3 means 2. Corresponding multipliers of x,y and z represents normal vector.

multiplying by 6, we get, 6(x-1)+3(y-2)+2(z-3)=0 or, 6x-6+3y-6+2z-6=o or, 6x+3y+2z=18

now, the normal vector will be (a,b,c) i.e, (6,3,2).

taking the lcm of 1,2 and 3 we need to multiply x-1 by 6 y-2 by 3 and z-3 by 2 . then the denominator is sent to rhs and we get standard equation of plane from where we can find the normal vector

• a(x-x0)+b(y-y0)+c(z-z0)=0
• where a=1
• b=1/2
• c=1/3
• x0=1 y0=2 z0=3
• so n=<1,1/2,1/3>
• linear dependent of n is <6,3,2>

Direction Ratio of given plane equal to DR's of normal of plane are 6,3,2