Find the vector equation of the plane which is at a distance of `6/sqrt(29)` from the origin and its normal vector from the origin is `2hati-3hatj+4hatk.` Also find its Cartesian equations

Find the vector equation of the plane which is at a distance of 6/sqrt29 from the origin and its normal vector from the origin is 2hati-3hatj+4hatk . Also find its cartesian form.

Find the vector equation of the plane which is at a distance of 6/sqrt29 from the origin and its normal vector from the origin is 2hati-3hatj+4hatk . Also find its cartesian form.

The vector equation of a plane which is at a distance of 6/sqrt29 from the origin and its normal vector from the origin is 2hati-3hatj+4hatk is

Find the vector equation of the plane which is at a distance of (6)/(sqrt(29)) the from the origin and its normal vector from the origin is 2hati-3hatj+4hatk . Also find its cartesian form.

Find the vector equation of the plane which is at a distance of 6/(sqrt(29)) from the origin and its normal vector from the origin is 2 hat i-3 hat j+4 hat k . Also find its cartesian form.

Find the vector equations of the plane which is at a distance of 7 from the origin and its normal vector from the origin is 2hati-6hatj+3hatk .

The vector equation of the plane which is at a distance of 3//sqrt(14) from the origin and the normal from the origin is 2hati-3hatj+hatk is

The vector equation of the plane which is at a distance (3)/(sqrt(14)) from the origin and the normal from the origin is 2hati + -3hatj + hatk is