Find the equation of the plane passing through the line of intersection of the planes `vecr.(hati + hatj + hatk) = 1` and `vecr.(2hati + 3hatj -hatk) + 4 = 0` and parallel to x-axis.
The given planes are
The equation of any plane passing through the line of intersection of these planes is
Its direction ratios are (2λ + 1), (3λ + 1), and (1 − λ).
The required plane is parallel to x-axis. Therefore, its normal is perpendicular to x-axis.
The direction ratios of x-axis are 1, 0, and 0.
Therefore, its Cartesian equation is y − 3z + 6 = 0
This is the equation of the required plane.