If x, y and z are non-zero real numbers and v | vedclass

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(C) Given, $\vec{a}=x \hat{i}+2 \hat{j}, \vec{b}=y \hat{j}+3 \hat{k}$ and $\vec{c}=x \hat{i}+y \hat{j}+z \hat{k}$.The cross product $\vec{a} 	imes \v

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$9$

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(C) Given, $\vec{a}=x \hat{i}+2 \hat{j}, \vec{b}=y \hat{j}+3 \hat{k}$ and $\vec{c}=x \hat{i}+y \hat{j}+z \hat{k}$.The cross product $\vec{a} 	imes \vec{b}$ is given by:$\vec{a} 	imes \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ x & 2 & 0 \ 0 & y & 3 \end{vmatrix} = \hat{i}(6-0) - \hat{j}(3x-0) + \hat{k}(xy-0) = 6\hat{i} - 3x\hat{j} + xy\hat{k}$.Comparing this with the given $\vec{a} 	imes \vec{b} = z\hat{i} - 3\hat{j} + \hat{k}$, we get:$6 = z$, $-3x = -3 \Rightarrow x = 1$, and $xy = 1$.Since $x=1$, we have $1 \cdot y = 1 \Rightarrow y = 1$.Thus, $\vec{a} = \hat{i} + 2\hat{j}$, $\vec{b} = \hat{j} + 3\hat{k}$, and $\vec{c} = \hat{i} + \hat{j} + 6\hat{k}$.The scalar triple product $[\vec{a} \vec{b} \vec{c}]$ is the determinant of the components of $\vec{a}, \vec{b}, \vec{c}$:$[\vec{a} \vec{b} \vec{c}] = \begin{vmatrix} 1 & 2 & 0 \ 0 & 1 & 3 \ 1 & 1 & 6 \end{vmatrix} = 1(6-3) - 2(0-3) + 0(0-1) = 3 + 6 = 9$.

If $x, y$ and $z$ are non-zero real numbers and $\vec{a}=x \hat{i}+2 \hat{j}, \vec{b}=y \hat{j}+3 \hat{k}$ and $\vec{c}=x \hat{i}+y \hat{j}+z \hat{k}$ are such that $\vec{a} \times \vec{b}=z \hat{i}-3 \hat{j}+\hat{k}$, then $[\vec{a} \vec{b} \vec{c}]$ equals to

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