For three vectors vec{A} = (-x hat{i} - 6 hat | vedclass

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(B) The scalar triple product $\vec{A} \cdot (\vec{B} 	imes \vec{C}) = 0$ implies that the three vectors are coplanar.This can be calculated using th

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$4$

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(B) The scalar triple product $\vec{A} \cdot (\vec{B} 	imes \vec{C}) = 0$ implies that the three vectors are coplanar.This can be calculated using the determinant of the components:$\begin{vmatrix} -x & -6 & -2 \ -1 & 4 & 3 \ -8 & -1 & 3 \end{vmatrix} = 0$Expanding the determinant along the first row:$-x(4(3) - 3(-1)) - (-6)((-1)(3) - 3(-8)) + (-2)((-1)(-1) - 4(-8)) = 0$$-x(12 + 3) + 6(-3 + 24) - 2(1 + 32) = 0$$-15x + 6(21) - 2(33) = 0$$-15x + 126 - 66 = 0$$-15x + 60 = 0$$15x = 60$$x = 4$

For three vectors $\vec{A} = (-x \hat{i} - 6 \hat{j} - 2 \hat{k})$,$\vec{B} = (-\hat{i} + 4 \hat{j} + 3 \hat{k})$ and $\vec{C} = (-8 \hat{i} - \hat{j} + 3 \hat{k})$,if $\vec{A} \cdot (\vec{B} \times \vec{C}) = 0$,then the value of $x$ is:

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