Consider the following planes $$P : x + y - 2z + 7 = 0$$, $$Q : x + y + 2z + 2 = 0$$, $$R : 3x + 3y - 6z - 11 = 0$$

For any plane written as $$ax + by + cz + d = 0$$, the vector $$\mathbf{n} = (a,\,b,\,c)$$ is called its normal vector.

• Two planes are parallel if and only if their normal vectors are parallel (scalar multiples).
• Two planes are perpendicular if and only if their normal vectors are perpendicular, i.e. their dot product is $$0$$.

Write the normal vectors of the three given planes:

Plane $$P: x + y - 2z + 7 = 0$$ has normal $$\mathbf{n}_P = (1,\,1,\,-2).$$

Plane $$Q: x + y + 2z + 2 = 0$$ has normal $$\mathbf{n}_Q = (1,\,1,\,2).$$

Plane $$R: 3x + 3y - 6z - 11 = 0$$ has normal $$\mathbf{n}_R = (3,\,3,\,-6).$$

Step 1 - Check for parallelism.

Compare $$\mathbf{n}_P$$ and $$\mathbf{n}_R$$: $$\mathbf{n}_R = 3\,(1,\,1,\,-2) = 3\,\mathbf{n}_P.$$ Since one is a scalar multiple of the other, $$\mathbf{n}_P \parallel \mathbf{n}_R$$, hence planes $$P$$ and $$R$$ are parallel.

For the other pairs:   • Between $$\mathbf{n}_P$$ and $$\mathbf{n}_Q$$, the component ratios are $$1/1 = 1$$, $$1/1 = 1$$, $$-2/2 = -1$$, not all equal ⇒ not parallel.   • Between $$\mathbf{n}_Q$$ and $$\mathbf{n}_R$$, the ratios $$3/1 = 3$$, $$3/1 = 3$$, $$-6/2 = -3$$ differ in sign ⇒ not parallel.

Step 2 - Check for perpendicularity using dot products.

$$\mathbf{n}_P \cdot \mathbf{n}_R = 1\cdot3 + 1\cdot3 + (-2)\cdot(-6) = 3 + 3 + 12 = 18 \neq 0,$$ so planes $$P$$ and $$R$$ are not perpendicular.

$$\mathbf{n}_Q \cdot \mathbf{n}_R = 1\cdot3 + 1\cdot3 + 2\cdot(-6) = 3 + 3 - 12 = -6 \neq 0,$$ so $$Q$$ and $$R$$ are not perpendicular.

$$\mathbf{n}_P \cdot \mathbf{n}_Q = 1\cdot1 + 1\cdot1 + (-2)\cdot2 = 1 + 1 - 4 = -2 \neq 0,$$ so $$P$$ and $$Q$$ are not perpendicular.

Only the statement “planes $$P$$ and $$R$$ are parallel” is true.

Option D which is: $$P$$ and $$R$$ are parallel