Let the lines frac{x-1}{lambda}=frac{y-2}{1}= | vedclass

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(D) For two lines to be coplanar,the determinant of the vector connecting points on the lines and the direction vectors must be zero.Given points $A(1

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$(0,4,5)$

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(D) For two lines to be coplanar,the determinant of the vector connecting points on the lines and the direction vectors must be zero.Given points $A(1, 2, 3)$ on $L_1$ and $B(-26, -18, -28)$ on $L_2$.The vector $\vec{AB} = (-26-1, -18-2, -28-3) = (-27, -20, -31)$.The condition for coplanarity is $\begin{vmatrix} -27 & -20 & -31 \ \lambda & 1 & 2 \ -2 & 3 & \lambda \end{vmatrix} = 0$.Expanding the determinant: $-27(\lambda - 6) + 20(\lambda^2 + 4) - 31(3\lambda + 2) = 0$.$-27\lambda + 162 + 20\lambda^2 + 80 - 93\lambda - 62 = 0 \Rightarrow 20\lambda^2 - 120\lambda + 180 = 0 \Rightarrow \lambda^2 - 6\lambda + 9 = 0 \Rightarrow (\lambda - 3)^2 = 0$.Thus,$\lambda = 3$.The normal vector $\vec{n}$ to the plane is the cross product of the direction vectors $\vec{v_1} = (3, 1, 2)$ and $\vec{v_2} = (-2, 3, 3)$.$\vec{n} = \vec{v_1} 	imes \vec{v_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 3 & 1 & 2 \ -2 & 3 & 3 \end{vmatrix} = \hat{i}(3-6) - \hat{j}(9+4) + \hat{k}(9+2) = -3\hat{i} - 13\hat{j} + 11\hat{k}$.The equation of the plane is $-3(x-1) - 13(y-2) + 11(z-3) = 0 \Rightarrow -3x + 3 - 13y + 26 + 11z - 33 = 0 \Rightarrow -3x - 13y + 11z - 4 = 0$,or $3x + 13y - 11z + 4 = 0$.Checking the points:For $(0, 4, 5): 3(0) + 13(4) - 11(5) + 4 = 52 - 55 + 4 = 1 
eq 0$.Thus,the point $(0, 4, 5)$ does not lie on the plane $P$.

Let the lines $\frac{x-1}{\lambda}=\frac{y-2}{1}=\frac{z-3}{2}$ and $\frac{x+26}{-2}=\frac{y+18}{3}=\frac{z+28}{\lambda}$ be coplanar and $P$ be the plane containing these two lines. Then which of the following points does $NOT$ lie on $P$?

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