Answer 1

We have,

`vecr.(hati + hatj + hatk) - 1 = 0`         ....(1)

`vecr.(2hati + 3hatj - hatk) + 4 = 0`     .....(2)

Equation of plane passing through the intersection of the planes (i) and (ii), is given by

`[vecr.(hati + hatj + hatk) - 1 ] = λ[vecr. (2hati + 3hatj - hatk) + 4 ] = 0`

⇒ `vecr.[(1 + 2λ)hati + (1 + 3λ)hatj + (1 - λ)hatk] - 1 + 4λ = 0`        ....(iii)

If plane (iii) is parallel to x-axis, then

1 + 2λ = 0 ⇒ λ = `-1/2`

Therefore, the equation of the required plane is `vecr.(-hatj + 3hatk) = 6`

Distance of the plane `vecr.(-hatj + 3hatk) = 6` from x-axis is given by,

`d = ((1 xx 0 - 3 xx 0 + 6)/(sqrt(1^2 + 3^2))) = 6/sqrt10`