If vec{a} = lambda x hat{i} + y hat{j} + 4z h | vedclass

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(C) Given the vectors $\vec{a} = (\lambda x) \hat{i} + (y) \hat{j} + (4z) \hat{k}$,$\vec{b} = (y) \hat{i} + (x) \hat{j} + (3y) \hat{k}$,and $\vec{c} =

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$2$

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(C) Given the vectors $\vec{a} = (\lambda x) \hat{i} + (y) \hat{j} + (4z) \hat{k}$,$\vec{b} = (y) \hat{i} + (x) \hat{j} + (3y) \hat{k}$,and $\vec{c} = (-z) \hat{i} + (-2z) \hat{j} - (\lambda + 1) \hat{k}$.Given the condition $\vec{a} + \vec{b} + \vec{c} = \vec{0}$,we sum the components:$i$-component: $\lambda x + y - z = 0$  $(1)$$j$-component: $y + x - 2z = 0$  $(2)$$k$-component: $4z + 3y - (\lambda + 1) = 0$  $(3)$From $(2)$,$x + y = 2z$. Substituting this into a system where $x, y, z$ are not all zero,we look for a non-trivial solution.For the system to have a non-trivial solution,the determinant of the coefficients must be zero.However,re-evaluating the vector sum $\vec{a} + \vec{b} = \vec{c}$ implies $\vec{c} = -(\vec{a} + \vec{b})$.Comparing components: $-z = \lambda x + y$,$-2z = y + x$,and $-(\lambda + 1) = 4z + 3y$.Solving the system leads to $\lambda = 2$.

List-$I$	List-$II$
$(A)$ $[\mathbf{a} \mathbf{b} \mathbf{c}]$	$1. \|\mathbf{a}\|\|\mathbf{b}\|\cos(\mathbf{a}, \mathbf{b})$
$(B)$ $(\mathbf{c} \times \mathbf{a}) \times \mathbf{b}$	$2. (\mathbf{a} \cdot \mathbf{c})\mathbf{b} - (\mathbf{a} \cdot \mathbf{b})\mathbf{c}$
$(C)$ $\mathbf{a} \times (\mathbf{b} \times \mathbf{c})$	$3. \mathbf{a} \cdot (\mathbf{b} \times \mathbf{c})$
$(D)$ $\mathbf{a} \cdot \mathbf{b}$	$4. \|\mathbf{a}\|\|\mathbf{b}\|$
	$5. (\mathbf{b} \cdot \mathbf{c})\mathbf{a} - (\mathbf{a} \cdot \mathbf{b})\mathbf{c}$

List $I$	List $II$
$P$. Volume of parallelepiped determined by vectors $\vec{a}, \vec{b}$ and $\vec{c}$ is $2$. Then the volume of the parallelepiped determined by vectors $2(\vec{a} \times \vec{b}), 3(\vec{b} \times \vec{c})$ and $(\vec{c} \times \vec{a})$ is	$1$. $100$
$Q$. Volume of parallelepiped determined by vectors $\vec{a}, \vec{b}$ and $\vec{c}$ is $5$. Then the volume of the parallelepiped determined by vectors $3(\vec{a}+\vec{b}), (\vec{b}+\vec{c})$ and $2(\vec{c}+\vec{a})$ is	$2$. $30$
$R$. Area of a triangle with adjacent sides determined by vectors $\vec{a}$ and $\vec{b}$ is $20$. Then the area of the triangle with adjacent sides determined by vectors $(2\vec{a}+3\vec{b})$ and $(\vec{a}-\vec{b})$ is	$3$. $24$
$S$. Area of a parallelogram with adjacent sides determined by vectors $\vec{a}$ and $\vec{b}$ is $30$. Then the area of the parallelogram with adjacent sides determined by vectors $(\vec{a}+\vec{b})$ and $\vec{a}$ is	$4$. $60$

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