Answer
See the detailed answer below.
Work Step by Step
a)
We know that
$$\vec C\times \vec D=0$$
where $\vec C=3\;\hat i$ and we need to find $\vec D$.
$$\vec C\times \vec D=(3\;\hat i+0\;\hat j+0\;\hat k)\times (D_x\;\hat i+D_y\;\hat j+D_z\;\hat k)=0$$
$$ (3\;\hat i\times D_x\;\hat i)+(3\;\hat i\times D_y\;\hat j)+(3\;\hat i\times D_z\;\hat k)=0$$
Recall that $\hat i\times \hat i=\hat j\times \hat j=\hat k\times \hat k=0$
$$ \overbrace{(3\;\hat i\times D_x\;\hat i)}^{0}+ \overbrace{(3\;\hat i\times D_y\;\hat j)}^{3D_y\;\hat k}+\overbrace{(3\;\hat i\times D_z\;\hat k)}^{-3D_z\;\hat j}=0$$
$$ 0\;\hat i\times \hat i+ 3D_y\;\hat k-3 D_z\;\hat j=0$$
This means that whether vector $\vec D=0$, or $\vec D=N\;\hat i$
where $N$ is any real number.
$$\boxed{\vec D=0,\;{\rm or }\;\vec D=N\;\hat i}$$
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b)
We know that
$$\vec C\times \vec E=6\;\hat k$$
$$3\;\hat i\times(E_x\;\hat i+E_y\;\hat j+E_z\;\hat k)=6\;\hat k$$
$$\overbrace{3\;\hat i\times E_x\;\hat i}^{0}+\overbrace{3\;\hat i\times E_y\;\hat j}^{3E_y\;\hat k}+\overbrace{3\;\hat i\times E_z\;\hat k}^{-3E_z\;\hat j}=6\;\hat k$$
It is obvious that $3E_y\;\hat k=6\;\hat k$ and hence
$$E_y=2$$
Also, $-3E_z\;\hat j=0$, and hence $E_z=0$.
Since $\hat i\times \hat i=0$, whether vector $\vec E=2\;\hat j$, or $\vec D=N\;\hat i+2\;\hat j$
where $N$ is any real number.
$$\boxed{\vec E=2\;\hat j,\;{\rm or }\;\vec E=N\;\hat i+2\;\hat j}$$
where $N$ is any real number.
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c) We know that
$$\vec C\times \vec F=-3\;\hat j$$
$$3\;\hat i\times(F_x\;\hat i+F_y\;\hat j+F_z\;\hat k)=-3\;\hat j$$
$$\overbrace{3\;\hat i\times F_x\;\hat i}^{0}+\overbrace{3\;\hat i\times F_y\;\hat j}^{3F_y\;\hat k}+\overbrace{3\;\hat i\times F_z\;\hat k}^{-3F_z\;\hat j}=6\;\hat j$$
Thus, $-3F_z=-3\;\hat j$ Hence, $F_z=1$
$$\boxed{\vec F=1\;\hat k,\;{\rm or }\;\vec F=N\;\hat i+1\;\hat k}$$
