Let a=p(hat{i}+hat{j}+hat{k}), b=hat{i}+hat{j | vedclass

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(D) The scalar triple product $[abc]$ is given by the determinant of the components of the vectors $a$, $b$, and $c$.$[abc] = \begin{vmatrix} p & p &

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$\left[\frac{-5}{3}, \frac{5}{9}\right]$

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(D) The scalar triple product $[abc]$ is given by the determinant of the components of the vectors $a$, $b$, and $c$.$[abc] = \begin{vmatrix} p & p & p \ 1 & 1 & -2 \ 2 & -1 & 2 \end{vmatrix}$Expanding the determinant along the first row:$[abc] = p(1(2) - (-2)(-1)) - p(1(2) - (-2)(2)) + p(1(-1) - 1(2))$$[abc] = p(2 - 2) - p(2 + 4) + p(-1 - 2)$$[abc] = p(0) - p(6) + p(-3) = -9p$Given that $-5 \leq [abc] \leq 15$, we substitute $[abc] = -9p$:$-5 \leq -9p \leq 15$Dividing by $-9$ and reversing the inequality signs:$\frac{15}{-9} \leq p \leq \frac{-5}{-9}$$\frac{-5}{3} \leq p \leq \frac{5}{9}$Thus, $p \in \left[\frac{-5}{3}, \frac{5}{9}ight]$.

Let $a=p(\hat{i}+\hat{j}+\hat{k})$, $b=\hat{i}+\hat{j}-2\hat{k}$, and $c=2\hat{i}-\hat{j}+2\hat{k}$ be three vectors. If the value of $[abc]$ is not more than $15$ and not less than $-5$, then $p$ lies in the interval:

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