Loop 1

14 - 2(I_{1} - I_{3}) - 1(I_{1} - I_{2})
= 0

therefore

3I_{1} - I_{2} -2I_{3} = 14[Eqn. 1]

Loop 2

1(I_{2} - I_{1}) + 1(I_{2} - I_{3})
+ 2I_{2} = 0

therefore

I_{1} -4I_{2} + I_{3} = 0 [Eqn. 2]

Loop 3

1(I_{3} - I_{2}) + 2(I_{3} - I_{1})
+ I_{3} = 0

therefore

2I_{1} + I_{2} - 4I_{3} = 0 [Eqn. 3]

Solve Eqns. 1, 2, 3 for the currents.

From [2] I_{1} = 4I_{2} - I_{3}, substitute
this in [1] and [2] and get

11I_{2} - 5I_{3} = 14 [Eqn. 4]

and

I_{2} = (2/3)I_{3} [Eqn. 5]

Solve [4] and [5], to get

I_{3} = 6 amp

I_{2} = 4 amp

I_{1} = 10 amp

Equivalent R = V/I where I is the total current out of the battery,
i.e., I_{1}

R = 14/10 = 1.4 W