The equation of the plane passing through the point (1, 1, 1) and perpendicular to the planes 2x + y - 2z = 5 and 3x - 6y - 2z = 7 is

3D Geometry PYQ — JEE Advanced 2017 · Shift 2 · Q25 with Solution

Let the required plane be perpendicular to both given planes.

The normal vectors of the given planes are: $\vec n_1=(2,1,-2)$ $\vec n_2=(3,-6,-2)$

If a plane is perpendicular to both these planes, then its normal vector must be perpendicular to both $\vec n_1$ and $\vec n_2$ .

Hence the normal vector of the required plane is: $\vec n=\vec n_1\times \vec n_2$

Compute the cross product:

\hat i & \hat j & \hat k \\ 2 & 1 & -2 \\ 3 & -6 & -2 \end{vmatrix}$$ $$\vec n=\hat i\begin{vmatrix}1 & -2 \\-6 & -2\end{vmatrix}-\hat j\begin{vmatrix}2 & -2\\3 & -2\end{vmatrix}+\hat k\begin{vmatrix}2 & 1\\3 & -6\end{vmatrix}$$ $$=\hat i[(1)(-2)-(-2)(-6)]-\hat j[(2)(-2)-(-2)(3)]+\hat k[(2)(-6)-(1)(3)]$$ $$=\hat i(-2-12)-\hat j(-4+6)+\hat k(-12-3)$$ $$=-14\hat i-2\hat j-15\hat k$$ Any scalar multiple is also valid, so we may take $$\vec n=(14,2,15)$$

Equation of plane through point $(1,1,1)$ with normal vector $(14,2,15)$ is: $14(x-1)+2(y-1)+15(z-1)=0$

Expanding: $14x-14+2y-2+15z-15=0$ $14x+2y+15z-31=0$ $14x+2y+15z=31$

Compare with options:

Therefore, the correct option is: $\boxed{\text{D}}$

Verification with stored correct answer:

Stored correct answer is D, which matches our derived answer.