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.

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#### Solution

The given planes are

The equation of any plane passing through the line of intersection of these planes is

r→.i⏜+j⏜+k⏜-1+λr→.2i⏜+3j⏜-k⏜+4=0

r→.2λ+1i⏜+3λ+1j⏜+1-λk⏜+4λ-1=0 ...(1)

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* − 3*z* + 6 = 0

This is the equation of the required plane.

Is there an error in this question or solution?

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