A parallelepiped is formed using three non collinear vectors whose magnitudes are 1, 2, 3. Angle between any of the vector with....

Question

A parallelepiped is formed using three non collinear vectors whose magnitudes are 1, 2, 3. Angle between any of the vector with normal of the plane determined by other two is π/3. Then the ratio of the surface area of the parallelogram to the volume of the parallelogram is?

My method:

Heres a slightly rough figure:

Now I assumed:

OA = $√3/2 i +j/2$

OB = $-2 i$

OC = $(3√3)/2 i -3k/2$

So volume equal to [OA OB OC] = 3/2

Area of face with OA, OB = mod (OA × OB) = 1

Area of face with OA, OC = mod (OA × OC) = (3√7)/4

Area of face with OB, OC = mod (OB × OC) = 3

So total surface area = 2 (1+3+(3√7)/4)

The ratio comes out to be something root 7

But given answer is 22/3. Where did i go wrong?