Q.The direction cosines of the normal to the plane $2x-y+2z=3$ are
(A) $\dfrac{2}{3}, \dfrac{-1}{3}, \dfrac{2}{3}$
(B) $\dfrac{-2}{3}, \dfrac{1}{3}, \dfrac{-2}{3}$
(C) $\dfrac{2}{3}, \dfrac{1}{3}, \dfrac{2}{3}$
(D) $\dfrac{2}{3}, \dfrac{-1}{3}, \dfrac{-2}{3}$

The cartesian plane $lx+my+nz=p$ has vector form $\vec r\cdot(l\hat i+m\hat j+n\hat k)=p$.

The plane $3x+4y-2z=5$ has normal direction ratios $(3,4,-2)$.

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Divide the plane equation by the magnitude of its normal vector, arranging so the constant on the right is positive.

Given plane: $x+2y-3z-6=0$, i.e. $x+2y-3z=6$.

The normal form of a plane is $lx+my+nz=p$ where $(l,m,n)$ are direction cosines of the normal (so $l^2+m^2+n^2=1$) and $p\ge 0$ is the perpendicular distance from the origin.

The normal vector to the given plane is $(1,2,-3)$. Its magnitude:

$\sqrt{1^{2}+2^{2}+(-3)^{2}} = \sqrt{1+4+9} = \sqrt{14}$

Divide both sides of $x+2y-3z=6$ by $\sqrt{14}$:

$\frac{1}{\sqrt{14}}x + \frac{2}{\sqrt{14}}y - \frac{3}{\sqrt{14}}z = \frac{6}{\sqrt{14}}$

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