If bar{a}=2 hat{i}-hat{j}+hat{k}, bar{b}=hat{ | vedclass

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(C) Since the vectors $\bar{a}, \bar{b},$ and $\bar{c}$ are coplanar,their scalar triple product must be zero,i.e.,$[\bar{a} \bar{b} \bar{c}] = 0$.Thi

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$x^2+3 x=4$

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(C) Since the vectors $\bar{a}, \bar{b},$ and $\bar{c}$ are coplanar,their scalar triple product must be zero,i.e.,$[\bar{a} \bar{b} \bar{c}] = 0$.This is equivalent to the determinant of the matrix formed by their components being zero:$\begin{vmatrix} 2 & -1 & 1 \ 1 & 2 & -3 \ 3 & \lambda & 5 \end{vmatrix} = 0$Expanding the determinant along the first row:$2(2 	imes 5 - (-3) 	imes \lambda) - (-1)(1 	imes 5 - (-3) 	imes 3) + 1(1 	imes \lambda - 2 	imes 3) = 0$$2(10 + 3\lambda) + 1(5 + 9) + 1(\lambda - 6) = 0$$20 + 6\lambda + 14 + \lambda - 6 = 0$$7\lambda + 28 = 0$$\lambda = -4$Now,check which equation has $\lambda = -4$ as a root:For option $C$: $x^2 + 3x = 4 \Rightarrow x^2 + 3x - 4 = 0 \Rightarrow (x+4)(x-1) = 0$.Thus,$x = -4$ is a root of the equation $x^2 + 3x = 4$.

If $\bar{a}=2 \hat{i}-\hat{j}+\hat{k}, \bar{b}=\hat{i}+2 \hat{j}-3 \hat{k}$ and $\bar{c}=3 \hat{i}+\lambda \hat{j}+5 \hat{k}$ are coplanar,then $\lambda$ is the root of the equation

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