A

To find three vectors that are perpendicular to the vector \( \vec{A} = 3\hat{i} - 4\hat{j} + 0\hat{k} \), we can apply the dot product property. A vector \( \vec{B} = a\hat{i} + b\hat{j} + c\hat{k} \) is perpendicular to \( \vec{A} \) if its dot product with \( \vec{A} \) equals zero:
\[ \vec{A} \cdot \vec{B} = 3a - 4b + 0c = 0 \]
Therefore, we need to solve \( 3a - 4b = 0 \) for various values of \( a \) and \( b \). This leads to: \( b = \frac{3}{4}a \).
We also need the magnitude of the resulting vector \( \vec{B} \) to be 10 units:
\[ \sqrt{a^2 + b^2 + c^2} = 10 \]
Substituting for \( b \):
\[ \sqrt{a^2 + \left(\frac{3}{4}a\right)^2 + c^2} = 10 \]
This simplifies to:
\[ \sqrt{a^2 + \frac{9}{16}a^2 + c^2} = 10 \]
\[ \sqrt{\frac{25}{16}a^2 + c^2} = 10 \]
Squaring both sides gives:
\[ \frac{25}{16}a^2 + c^2 = 100 \]
Choose \( a = 4 \):
\[ \frac{25}{16}(4)^2 + c^2 = 100 \]
\[ 25 + c^2 = 100 \]
\[ c^2 = 75 \Rightarrow c = 5\sqrt{3} \]
Therefore, one solution could be \( \vec{B_1} = 4\hat{i} + 3\hat{j} + 5\sqrt{3}\hat{k} \).
By similarly choosing different values for \( a \) and \( c \), we can find additional perpendicular vectors. Thus, a total of three vectors can be derived by varying \( a \), and we can get:\
1. \( 4\hat{i} + 3\hat{j} + 5\sqrt{3}\hat{k} \)
2. \( -4\hat{i} + 3\hat{j} + 5\sqrt{3}\hat{k} \)
3. \( 0\hat{i} + (-4)\hat{j} + 10\hat{k} \)
Therefore, several perpendicular vectors can be found with defined relationships and magnitudes.