If L_1 and L_2 are two lines which pass throu | vedclass

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(C) Since the lines $L_1$ and $L_2$ pass through the origin $(0, 0, 0)$,the plane containing them also passes through the origin. The equation of such

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$2x - y + z = 0$

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(C) Since the lines $L_1$ and $L_2$ pass through the origin $(0, 0, 0)$,the plane containing them also passes through the origin. The equation of such a plane is given by $ax + by + cz = 0$.The normal vector $\vec{n}$ to the plane is the cross product of the direction vectors of the two lines,$\vec{v}_1 = 3\hat{i} + \hat{j} - 5\hat{k}$ and $\vec{v}_2 = 2\hat{i} + 3\hat{j} - \hat{k}$.$\vec{n} = \vec{v}_1 	imes \vec{v}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 3 & 1 & -5 \ 2 & 3 & -1 \end{vmatrix}$$\vec{n} = \hat{i}(-1 - (-15)) - \hat{j}(-3 - (-10)) + \hat{k}(9 - 2)$$\vec{n} = 14\hat{i} - 7\hat{j} + 7\hat{k}$Dividing by $7$,we get the normal vector as $2\hat{i} - \hat{j} + \hat{k}$.Thus,the equation of the plane is $2(x - 0) - 1(y - 0) + 1(z - 0) = 0$,which simplifies to $2x - y + z = 0$.

If $L_1$ and $L_2$ are two lines which pass through the origin and have direction ratios $(3, 1, -5)$ and $(2, 3, -1)$ respectively,then the equation of the plane containing $L_1$ and $L_2$ is

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