Answer:

Solenoidal: A vector $\overline F$ is called as solonoidal if divergence of $\overline F$ is zero. i.e. $\nabla.\overline F=0$ 

We have given

$\overline{F}= \dfrac {{\overline {a}} {\times {\overline {r}}}} {r^n} \cdots\cdots(1)$

We know that,

$\nabla=\big(\dfrac \partial {\partial{x}}\hat i+\dfrac \partial {\partial{y}}\hat j+\dfrac \partial {\partial{z}}\hat k) \cdots\cdots(2)$

$\overline a=a_1 \hat{i}+a_2 \hat{j}+a_3 \hat{k}$

$\overline r=x \hat{i}+y \hat{j}+z \hat{k}$

$ |r|= \sqrt{x^2+y^2+z^2} $

$r^2=x^2+y^2+z^2 \cdots\cdots(3)$

${\overline {a}} {\times {\overline {r}}}= \left[\begin{matrix} \hat i &\hat j & \hat k \\ a_1 & a_2 & a_3 \\ x & y & z \end{matrix}\right]$

        $=\hat i[(a_2.z)-(a_3.y)]-\hat j[(a_1.z)-(a_3.x)]+\hat k[(a_1.y)-(a_2.x)] \cdots\cdots(4)$

$\overline{F}= \dfrac {{\overline {a}} {\times {\overline {r}}}} {r^n}$

Now partially differentiate eqn (3) with respect to x

  $\dfrac \partial {\partial{x}} {r^2}=\big(\dfrac \partial {\partial{x}}) . {\big (x^2+y^2 +z^2 \big)}$

 Take partial derivative of r with respect to x then again take derivative of r with respect to x because it r instead x.

$2r \dfrac {\partial r}{\partial x}=2x$    ∵ r is consist of x,y,z so it is not a constant.

    $\dfrac {\partial r}{\partial{x}}=\dfrac {x}{r}$

Similarly we get,

   $\dfrac {\partial r}{\partial{y}}=\dfrac {y}{r}$

Similarly we get,

   $\dfrac {\partial r}{\partial{z}}=\dfrac {z}{r}$

Putting eqn(3) & eqn(4) in eqn(1) we get,

$\therefore \overline F={\dfrac{\hat i[(a_2.z)-(a_3.y)]-\hat j[(a_1.z)-(a_3.x)]+\hat k[(a_1.y)-(a_2.x)]}{ r^n}}$

$\nabla.\overline F=\nabla.{\dfrac{\hat i[(a_2.z)-(a_3.y)]-\hat j[(a_1.z)-(a_3.x)]+\hat k[(a_1.y)-(a_2.x)]}{ r^n}}$

         $ =\big(\frac \partial {\partial{x}}\hat i+\frac \partial {\partial{y}}\hat j+\frac \partial {\partial{z}}\hat k).\Bigg({\dfrac{\hat i[(a_2.z)-(a_3.y)]+\hat j[(a_1.z)-(a_3.x)]+\hat k[(a_1.y)-(a_2.x)]}{ r^n}}\Bigg)$

         $ =\big(\frac \partial {\partial{x}}\hat i+\frac \partial {\partial{y}}\hat j+\frac \partial {\partial{z}}\hat k). \Bigg( {\dfrac {[(a_2.z)-(a_3.y)]} { r^n} \hat i} -{\dfrac {[(a_1.z)-(a_3.x)]} { r^n}} \hat j +{\dfrac {[(a_1.y)-(a_2.x)]}{ r^n}}\hat k \Bigg) $

         $ = \Bigg( {\dfrac \partial {\partial{x}}} {\dfrac {[(a_2.z)-(a_3.y)]} {r^n}} -{\dfrac \partial {\partial{y}}} {\dfrac {[(a_1.z)-(a_3.x)]} {r^n}} +{\dfrac \partial {\partial{z}}} {\dfrac {[(a_1.y)-(a_2.x)]} {r^n}} \Bigg) \cdots\cdots(5)$

First we find $ {\dfrac \partial {\partial{x}}} {\dfrac {[(a_2.z)-(a_3.y)]} {r^n}}$

$\therefore {\dfrac \partial {\partial{x}}} {\dfrac {[(a_2.z)-(a_3.y)]} {r^n}} = \dfrac {{{{r^n}. {\dfrac \partial {\partial{x}}} {[(a_2.z)-(a_3.y)]} - {[(a_2.z)-(a_3.y)]}. {\dfrac \partial {\partial{x}}}{r^n}}}}{(r^n)^2}$      $\because {\dfrac {\partial} {\partial{x}}} \dfrac {u} {v}= \dfrac {{{{v}. {\dfrac \partial {\partial{x}}} {u} - {u}. {\dfrac \partial {\partial{x}}}{v}}}}{v^2 }$

$=\dfrac {{{{r^n}. (0) - {[(a_2.z)-(a_3.y)]}.n.r^{n-1} {\dfrac {\partial {r}}{\partial{x}}}}}}{r^{2n}}$

$=\dfrac {- {[(a_2.z)-(a_3.y)]}.n.r^{n-1} {\dfrac {x}{r}}} {r^{2n}} \cdots\cdots \phi_1$

Now we find  $ {\dfrac \partial {\partial{y}}} {\dfrac {[(a_1.z)-(a_3.x)]} {r^n}} $

$\therefore {\dfrac \partial {\partial{y}}} {\dfrac {[(a_1.z)-(a_3.x)]} {r^n}} =\dfrac {{{{r^n}. {\dfrac \partial {\partial{y}}} {[(a_1.z)-(a_3.x)]} - {[(a_1.z)-(a_3.x)]}. {\dfrac \partial {\partial{y}}} {r^n}}}} {(r^n)^2}$

$=\dfrac {{{{r^n}. (0) - {[(a_1.z)-(a_3.x)]}.n.r^{n-1} {\dfrac {\partial {r}}{\partial{y}}}}}}{r^{2n}}$

$=\dfrac {- {[(a_1.z)-(a_3.x)]}.n.r^{n-1} {\dfrac {y}{r}}} {r^{2n}} \cdots\cdots \phi_2$

Now we find  ${\dfrac \partial {\partial{z}}} {\dfrac {[(a_1.y)-(a_2.x)]} {r^n}}$

$\therefore {\dfrac \partial {\partial{z}}} {\dfrac {[(a_1.y)-(a_2.x)]} {r^n}} =\dfrac {{{{r^n}. {\dfrac \partial {\partial{z}}} {[(a_1.y)-(a_2.x)]} - {[(a_1.y)-(a_2.x)]}. {\dfrac \partial {\partial{z}}} {r^n}}}} {(r^n)^2}$

$=\dfrac {{{{r^n}. (0) - {[(a_1.y)-(a_2.x)]}.n.r^{n-1} {\dfrac {\partial {r}}{\partial{z}}}}}}{r^{2n}}$

$=\dfrac {- {[(a_1.y)-(a_2.x)]}.n.r^{n-1} {\dfrac {z}{r}}} {r^{2n}} \cdots\cdots \phi_3$

Putting values of $\Phi_1,\Phi_2,\Phi_3$ in eqn (5) we get,

$\nabla.\overline F = \Bigg( \dfrac {- {[(a_2.z)-(a_3.y)]}.n.r^{n-1} {\dfrac {x}{r}}} {r^{2n}} \bigg) -\bigg(\dfrac {- {[(a_1.z)-(a_3.x)]}.n.r^{n-1} {\dfrac {y}{r}}} {r^{2n}} \bigg) +\bigg(\dfrac {- {[(a_1.y)-(a_2.x)]}.n.r^{n-1} {\dfrac {z}{r}}} {r^{2n}} \Bigg)$

         $= {-n.r^{n-1} }{\dfrac {1}{r}} \Bigg( \dfrac {{[(a_2.z)-(a_3.y)]}{x} - {[(a_1.z)-(a_3.x)]}{y} + {[(a_1.y)-(a_2.x)]}{z}} {r^{2n}} \Bigg)$

         $= {-n.r^{n-1} }{\dfrac {1}{r}} \Bigg( \dfrac {{[(a_2.xz)-(a_3.xy)]} - {[(a_1.yz)-(a_3.xy)]} + {[(a_1.yz)-(a_2.xz)]}} {r^{2n}} \Bigg)$

         $= {-n.r^{n-1} }{\dfrac {1}{r}} \Bigg( \dfrac {{(a_2.xz)-(a_3.xy)} - {(a_1.yz)+(a_3.xy)} + {(a_1.yz)-(a_2.xz)}} {r^{2n}} \Bigg)$

         $= {-n.r^{n-1} }{\dfrac {1}{r}} \Big( 0 \Big)$

$\nabla.\overline F=0$

 ∴ Given $\overline F$ is Solenoidal.