If the vectors phat{i} + hat{j} + hat{k},hat{ | vedclass

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(A) If three vectors $\vec{a}, \vec{b}, \vec{c}$ are coplanar,their scalar triple product is zero,i.e.,$[\vec{a} \vec{b} \vec{c}] = 0$.In determinant

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$-2$

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(A) If three vectors $\vec{a}, \vec{b}, \vec{c}$ are coplanar,their scalar triple product is zero,i.e.,$[\vec{a} \vec{b} \vec{c}] = 0$.In determinant form:$\begin{vmatrix} p & 1 & 1 \ 1 & q & 1 \ 1 & 1 & r \end{vmatrix} = 0$Applying $R_1 	o R_1 - R_3$ and $R_2 	o R_2 - R_3$:$\begin{vmatrix} p-1 & 0 & 1-r \ 0 & q-1 & 1-r \ 1 & 1 & r \end{vmatrix} = 0$Taking $(p-1)$ common from $R_1$ and $(q-1)$ common from $R_2$:$(p-1)(q-1) \begin{vmatrix} 1 & 0 & \frac{1-r}{p-1} \ 0 & 1 & \frac{1-r}{q-1} \ 1 & 1 & r \end{vmatrix} = 0$Expanding the determinant:$(p-1)(q-1) [1(r - \frac{1-r}{q-1}) - 0 + \frac{1-r}{p-1}(0 - 1)] = 0$$(p-1)(q-1) [\frac{rq - r - 1 + r}{q-1} - \frac{1-r}{p-1}] = 0$$(p-1)(rq - 1) - (q-1)(1-r) = 0$$pqr - p - rq + 1 - (q - qr - 1 + r) = 0$$pqr - p - q - r + 2 = 0$Therefore,$pqr - (p + q + r) = -2$.

If the vectors $p\hat{i} + \hat{j} + \hat{k}$,$\hat{i} + q\hat{j} + \hat{k}$,and $\hat{i} + \hat{j} + r\hat{k}$ $(p \neq 1, q \neq 1, r \neq 1)$ are coplanar,then find the value of $pqr - (p + q + r)$.

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