Let the acute angle bisector of the two planes $$x - 2y - 2z + 1 = 0$$ and $$2x - 3y - 6z + 1 = 0$$ be the plane P. Then which of the following points lies on P?

We have two given planes

$$\pi_1: \; x-2y-2z+1=0,$$

$$\pi_2: \; 2x-3y-6z+1=0.$$

For two planes $$a_1x+b_1y+c_1z+d_1=0$$ and $$a_2x+b_2y+c_2z+d_2=0,$$ the locus of points which are equidistant from the two planes is obtained from the standard formula

$$\frac{a_1x+b_1y+c_1z+d_1}{\sqrt{a_1^{2}+b_1^{2}+c_1^{2}}}= \pm\; \frac{a_2x+b_2y+c_2z+d_2}{\sqrt{a_2^{2}+b_2^{2}+c_2^{2}}}.$$

The ‘$$\pm$$’ gives the two bisector planes; one of them bisects the acute angle and the other bisects the obtuse angle between the given planes.

First we compute the magnitudes of the normal vectors of the given planes.

The normal to $$\pi_1$$ is $$\mathbf n_1=(1,-2,-2),$$ so

$$|\mathbf n_1|=\sqrt{1^{2}+(-2)^{2}+(-2)^{2}}=\sqrt{1+4+4}=3.$$

The normal to $$\pi_2$$ is $$\mathbf n_2=(2,-3,-6),$$ hence

$$|\mathbf n_2|=\sqrt{2^{2}+(-3)^{2}+(-6)^{2}}=\sqrt{4+9+36}=7.$$

We now substitute these values in the formula.

$$\frac{x-2y-2z+1}{3}=\pm\frac{2x-3y-6z+1}{7}.$$

Cross-multiplying gives the two possible bisector planes:

For the ‘$$+$$’ sign

$$7\,(x-2y-2z+1)-3\,(2x-3y-6z+1)=0,$$

and for the ‘$$-$$’ sign

$$7\,(x-2y-2z+1)+3\,(2x-3y-6z+1)=0.$$

Let us simplify each expression step by step.

Expanding the first (with the ‘$$+$$’ sign between the two fractions):

$$7x-14y-14z+7-6x+9y+18z-3=0,$$

so

$$x-5y+4z+4=0.$$ We denote this plane by $$\Pi_a.$$

Expanding the second (with the ‘$$-$$’ sign between the two fractions):

$$7x-14y-14z+7+6x-9y-18z+3=0,$$

which reduces to

$$13x-23y-32z+10=0.$$ Call this plane $$\Pi_b.$$

Exactly one of these two is the acute-angle bisector. To pick the correct one we examine the angle made by each candidate normal with the normals of the given planes.

The normal of $$\Pi_a$$ is $$\mathbf n_a=(1,-5,4).$$ Its dot product with $$\mathbf n_1$$ is

$$\mathbf n_a\!\cdot\!\mathbf n_1=1\cdot1+(-5)(-2)+4(-2)=1+10-8=3 \;(\gt0),$$

but with $$\mathbf n_2$$ we find

$$\mathbf n_a\!\cdot\!\mathbf n_2=1\cdot2+(-5)(-3)+4(-6)=2+15-24=-7 \;(\lt0).$$

The negative result means $$\Pi_a$$ makes an obtuse angle with $$\pi_2,$$ so $$\Pi_a$$ is the obtuse-angle bisector.

The normal of $$\Pi_b$$ is $$\mathbf n_b=(13,-23,-32).$$ Now

$$\mathbf n_b\!\cdot\!\mathbf n_1=13\cdot1+(-23)(-2)+(-32)(-2)=13+46+64=123\;(\gt0),$$

and

$$\mathbf n_b\!\cdot\!\mathbf n_2=13\cdot2+(-23)(-3)+(-32)(-6)=26+69+192=287\;(\gt0).$$

Since the dot product is positive with both normals, the angles between $$\Pi_b$$ and each of the given planes are acute. Hence

$$\,13x-23y-32z+10=0\,$$

is the required acute-angle bisector plane $$P.$$

Now we test each option. We substitute the coordinates into $$13x-23y-32z+10.$$

Option A: $$(0,2,-4)$$ gives $$13(0)-23(2)-32(-4)+10=0-46+128+10=92\neq0.$$

Option B: $$(-2,0,-\tfrac12)$$ gives $$13(-2)-23(0)-32\!\Bigl(-\tfrac12\Bigr)+10=-26+0+16+10=0.$$

Option C: $$(4,0,-2)$$ gives $$13(4)-23(0)-32(-2)+10=52+0+64+10=126\neq0.$$

Option D: $$(3,1,-\tfrac12)$$ gives $$13(3)-23(1)-32\!\Bigl(-\tfrac12\Bigr)+10=39-23+16+10=42\neq0.$$

Only Option B makes the left-hand side zero, so only that point lies on the acute angle bisector plane.

Hence, the correct answer is Option B.