Fdtxjr

Given that $veca$ is a constant vector and $vecr$ is a position vector.

We are asked to prove the following:
$$nablatimes(vecatimesvecr)=2veca$$
I tried two ways. Could prove it from one method but from another I am getting another result.

$$rule20cm0.4pt$$

1st method(with correct result):

$$LHS=nablatimes(vecatimesvecr)$$
Let $veca=a_1hati+a_2hatj+a_3hatk ,,$ and $,,vecr=xhati+yhatj+zhatk,$ where $a_1,a_2,a_3$ are constants and $x,y,z$ is any arbitrary point.

$$Simplifying,,,, (vecatimesvecr)$$

$$=beginvmatrixhati & hatj & hatk \ a_1 & a_2 & a_3 \ x & y & zendvmatrix$$
$$=hati(a_2z-a_3y)+hatj(a_3x-a_1z)+hatk(a_1y-a_2x)$$

Now performing the curl operation on the simplified result we obtained.
$$=nablatimes(simplified,result)$$
$$=beginvmatrixhati & hatj & hatk \ fracpartialpartial x &fracpartialpartial y & fracpartialpartial z \ a_2z-a_3y & a_3x-a_1z & a_1y-a_2xendvmatrix$$

$$=2a_1xhati+2a_2yhatj+2a_3zhatk$$
$$=2veca$$
$$=RHS$$
$$rule20cm0.4pt$$
2nd method with incorrect result.

We know that $$nablatimes(vecatimesvecr)=2[(nabla.vecr)veca-(nabla.veca)vecr]$$

Solving RHS,

$$Also,nabla.vecr=3,and, nabla.veca=0$$
$$=2[3veca-0vecr]$$
$$=6veca$$
which is incorrect. Where did I make mistake in this method?

The "vector calculus identity" you use in your second calculation is just false. You actually have
beginalign*
nabla times (veca times vecr) &= veca(nabla cdot vecr) - vecr(nabla cdot veca) + (vecr cdot nabla)veca - (veca cdot nabla) vecr \
&=
veca(nabla cdot vecr) - (veca cdot nabla) vecr.
endalign*
As you say, the first term on the RHS is then $3 veca$, but we need to calculate the other term. It is just the directional derivative of $vecr$ in the direction of $veca$ multiplied by the magnitude of $veca$, so this is just $veca$. Thus
$$
nabla times (veca times vecr) = 3 veca - veca = 2 veca,
$$
as desired.