If the points 2a+3b-c, a-2b+3c, 3a+lambda b-2 | vedclass

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(C) Let the points be $P_1 = 2a+3b-c, P_2 = a-2b+3c, P_3 = 3a+\lambda b-2c$ and $P_4 = a-6b+6c$.The vectors $\vec{P_1P_2}, \vec{P_1P_3}$ and $\vec{P_1

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$\frac{4}{9}, \frac{-8}{9}, \frac{1}{9}$

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(C) Let the points be $P_1 = 2a+3b-c, P_2 = a-2b+3c, P_3 = 3a+\lambda b-2c$ and $P_4 = a-6b+6c$.The vectors $\vec{P_1P_2}, \vec{P_1P_3}$ and $\vec{P_1P_4}$ are coplanar if their scalar triple product is zero.$\vec{P_1P_2} = (a-2b+3c) - (2a+3b-c) = -a-5b+4c$$\vec{P_1P_3} = (3a+\lambda b-2c) - (2a+3b-c) = a+(\lambda-3)b-c$$\vec{P_1P_4} = (a-6b+6c) - (2a+3b-c) = -a-9b+7c$The condition for coplanarity is $\begin{vmatrix} -1 & -5 & 4 \ 1 & \lambda-3 & -1 \ -1 & -9 & 7 \end{vmatrix} = 0$.Expanding the determinant: $-1(7(\lambda-3) - 9) + 5(7-1) + 4(-9 + (\lambda-3)) = 0$.$-1(7\lambda - 21 - 9) + 5(6) + 4(\lambda - 12) = 0$.$-7\lambda + 30 + 30 + 4\lambda - 48 = 0$.$-3\lambda + 12 = 0 \Rightarrow \lambda = 4$.The vector is $4\hat{i} - 8\hat{j} + \hat{k}$.The magnitude is $\sqrt{4^2 + (-8)^2 + 1^2} = \sqrt{16 + 64 + 1} = \sqrt{81} = 9$.The direction cosines are $\frac{4}{9}, \frac{-8}{9}, \frac{1}{9}$.

If the points $2a+3b-c, a-2b+3c, 3a+\lambda b-2c$ and $a-6b+6c$ are coplanar, then the direction cosines of the vector $\lambda \hat{i}-2\lambda \hat{j}+\hat{k}$ are

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