The two given forces,
[tex]\mathbf F_1 = (-4 \, \mathbf i + 7 \, \mathbf j) \, \mathrm N[/tex]
[tex]\mathbf F_2 = (2a \, \mathbf i - a \, \mathbf j) \, \mathrm N[/tex]
have resultant
[tex]\mathbf R = \mathbf F_1 + \mathbf F_2 = ((2a-4) \, \mathbf i + (7 - a) \, \mathbf j) \, \mathrm N[/tex]
R is parallel to the vector 3i + j, which means the angle between these two vectors is either θ = 0° or θ = 180°. Then
[tex]\mathbf R \cdot (3 \, \mathbf i + \mathbf j) = \|\mathbf R\| \|3\,\mathbf i+\mathbf j\| \cos(\theta)[/tex]
[tex]\implies 3(2a-4)+(7-a) = \pm \sqrt{10} \, \sqrt{(2a-4)^2+(7-a)^2}[/tex]
[tex]\implies 5a - 5 = \pm \sqrt{50} \, \sqrt{a^2 - 6a + 13}[/tex]
[tex]\implies a - 1 = \pm \sqrt2 \, \sqrt{a^2 - 6a + 13}[/tex]
Solve for a :
[tex]a - 1 = \pm \sqrt2 \, \sqrt{a^2 - 6a + 13}[/tex]
[tex]\implies (a - 1)^2 = \left(\pm \sqrt2 \, \sqrt{a^2-6a+13}\right)^2[/tex]
[tex]\implies a^2 - 2a + 1 = 2a^2 - 12a + 26[/tex]
[tex]\implies 0 = a^2 - 10a + 25[/tex]
[tex]\implies 0 = (a - 5)^2[/tex]
[tex]\implies \boxed{a = 5}[/tex]
