Find a vector v⃗ parallel to the intersection of the planes 3y−x−6z=2 and 2x+y+4z=7 .

To find a vector parallel to the intersection of the planes, we need to find the cross product of the normal vectors of the two planes.

The normal vector of the first plane, 3y - x - 6z = 2, is (-1, 3, -6).

The normal vector of the second plane, 2x + y + 4z = 7, is (2, 1, 4).

The cross product of these two vectors will give us a vector that is perpendicular to both of these vectors, and hence parallel to the line of intersection of the two planes.

The cross product of two vectors (a1, a2, a3) and (b1, b2, b3) is given by:

(a2b3 - a3b2, a3b1 - a1b3, a1b2 - a2b1)

So, the cross product of (-1, 3, -6) and (2, 1, 4) is:

(34 - (-6)1, -62 - (-1)4, -11 - 32) = (12 + 6, -12 + 4, -1 - 6) = (18, -8, -7)

So, the vector v that is parallel to the intersection of the planes is (18, -8, -7).