4. A particle \(P\) of mass 0.5 kg is in equilibrium under the action of three forces \(\mathbf { F } _ { 1 } , \mathbf { F } _ { 2 }\) and \(\mathbf { F } _ { 3 }\). $$\mathbf { F } _ { 1 } = ( 9 \mathbf { i } + 6 \mathbf { j } - 12 \mathbf { k } ) \mathbf { N } \quad \text { and } \quad \mathbf { F } _ { 2 } = ( 6 \mathbf { i } - 7 \mathbf { j } + 3 \mathbf { k } ) \mathbf { N } .$$

1. Find the force \(\mathbf { F } _ { 3 }\).
2. Forces \(\mathbf { F } _ { 2 }\) and \(\mathbf { F } _ { 3 }\) are removed so that \(P\) moves in a straight line \(A B\) under the action of the single force \(\mathbf { F } _ { 1 }\). The points \(A\) and \(B\) have position vectors ( \(2 \mathbf { i } - 9 \mathbf { j } + 7 \mathbf { k }\) )m and \(( 8 \mathbf { i } - 5 \mathbf { j } - \mathbf { k } ) \mathrm { m }\) respectively. The particle \(P\) is initially at rest at \(A\).
   1. Verify that \(\mathbf { F } _ { 1 }\) acts parallel to the vector \(\mathbf { A B }\).
   2. Find the work done by the force \(\mathbf { F } _ { 1 }\) as the particle moves from \(A\) to \(B\).
   3. By using the work-energy principle, find the speed of \(P\) as it reaches \(B\).