If the volume of a tetrahedron having bar{i}+ | vedclass

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(A) The volume of a tetrahedron with coterminous edges $\vec{a}, \vec{b}, \vec{c}$ is given by $V = \frac{1}{6} |[\vec{a} \vec{b} \vec{c}]|$.Given $\v

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$x^2+4 x-12=0$

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(A) The volume of a tetrahedron with coterminous edges $\vec{a}, \vec{b}, \vec{c}$ is given by $V = \frac{1}{6} |[\vec{a} \vec{b} \vec{c}]|$.Given $\vec{a} = \bar{i}+2 \bar{j}-3 \bar{k}$, $\vec{b} = 2 \bar{i}+\bar{j}-3 \bar{k}$, and $\vec{c} = 3 \bar{i}-\bar{j}+p \bar{k}$.The scalar triple product is $[\vec{a} \vec{b} \vec{c}] = \begin{vmatrix} 1 & 2 & -3 \ 2 & 1 & -3 \ 3 & -1 & p \end{vmatrix}$.Expanding the determinant: $1(p-3) - 2(2p+9) - 3(-2-3) = p-3 - 4p-18 + 15 = -3p-6$.Given volume $V = 2$, so $\frac{1}{6} |-3p-6| = 2$.$|-3p-6| = 12$.This implies $-3p-6 = 12$ or $-3p-6 = -12$.Case $1$: $-3p = 18 \implies p = -6$.Case $2$: $-3p = -6 \implies p = 2$.The values of $p$ are $2$ and $-6$.The quadratic equation whose roots are $2$ and $-6$ is $(x-2)(x+6) = 0$.$x^2+6x-2x-12 = 0 \implies x^2+4x-12 = 0$.

If the volume of a tetrahedron having $\bar{i}+2 \bar{j}-3 \bar{k}$, $2 \bar{i}+\bar{j}-3 \bar{k}$, and $3 \bar{i}-\bar{j}+p \bar{k}$ as its coterminous edges is $2$, then the values of $p$ are the roots of the equation

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