If the straight lines frac{x-1}{2}=frac{y+1}{ | vedclass

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(B) For two lines to be coplanar,the scalar triple product of the vector connecting a point on each line and the direction vectors of the lines must b

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$(B, C)$

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(B) For two lines to be coplanar,the scalar triple product of the vector connecting a point on each line and the direction vectors of the lines must be zero: $[\vec{a}-\vec{c}, \vec{b}, \vec{d}] = 0$.Given points $\vec{a} = (1, -1, 0)$ and $\vec{c} = (-1, -1, 0)$,the vector $\vec{a}-\vec{c} = (2, 0, 0)$.The direction vectors are $\vec{b} = 2\hat{i} + k\hat{j} + 2\hat{k}$ and $\vec{d} = 5\hat{i} + 2\hat{j} + k\hat{k}$.Setting the determinant to zero:$\left|\begin{array}{lll} 2 & 0 & 0 \ 2 & k & 2 \ 5 & 2 & k \end{array}ight| = 0 \Rightarrow 2(k^2 - 4) = 0 \Rightarrow k^2 = 4 \Rightarrow k = \pm 2$.Case $1$: If $k = 2$,the direction vectors are $\vec{b} = (2, 2, 2)$ and $\vec{d} = (5, 2, 2)$.The normal vector is $\vec{n}_1 = \vec{b} 	imes \vec{d} = \left|\begin{array}{lll} \hat{i} & \hat{j} & \hat{k} \ 2 & 2 & 2 \ 5 & 2 & 2 \end{array}ight| = 0\hat{i} + 6\hat{j} - 6\hat{k}$.The plane equation is $(\vec{r} - \vec{a}) \cdot \vec{n}_1 = 0 \Rightarrow 0(x-1) + 6(y+1) - 6(z-0) = 0 \Rightarrow y - z = -1$.Case $2$: If $k = -2$,the direction vectors are $\vec{b} = (2, -2, 2)$ and $\vec{d} = (5, 2, -2)$.The normal vector is $\vec{n}_2 = \vec{b} 	imes \vec{d} = \left|\begin{array}{lll} \hat{i} & \hat{j} & \hat{k} \ 2 & -2 & 2 \ 5 & 2 & -2 \end{array}ight| = 0\hat{i} + 14\hat{j} + 14\hat{k}$.The plane equation is $(\vec{r} - \vec{a}) \cdot \vec{n}_2 = 0 \Rightarrow 0(x-1) + 14(y+1) + 14(z-0) = 0 \Rightarrow y + z = -1$.Thus,the planes are $y-z=-1$ and $y+z=-1$,which correspond to options $(C)$ and $(B)$.

If the straight lines $\frac{x-1}{2}=\frac{y+1}{k}=\frac{z}{2}$ and $\frac{x+1}{5}=\frac{y+1}{2}=\frac{z}{k}$ are coplanar,then the plane$(s)$ containing these two lines is(are):

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