Let the planes be 3x - 6y - 2z = 15 and 2x + | vedclass

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(D) The normal vectors to the planes are $\vec{n}_1 = 3\hat{i} - 6\hat{j} - 2\hat{k}$ and $\vec{n}_2 = 2\hat{i} + \hat{j} - 2\hat{k}$.The direction ve

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Statement-$1$ is false,Statement-$2$ is true.

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(D) The normal vectors to the planes are $\vec{n}_1 = 3\hat{i} - 6\hat{j} - 2\hat{k}$ and $\vec{n}_2 = 2\hat{i} + \hat{j} - 2\hat{k}$.The direction vector $\vec{v}$ of the line of intersection is given by $\vec{v} = \vec{n}_1 	imes \vec{n}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 3 & -6 & -2 \ 2 & 1 & -2 \end{vmatrix} = \hat{i}(12 - (-2)) - \hat{j}(-6 - (-4)) + \hat{k}(3 - (-12)) = 14\hat{i} + 2\hat{j} + 15\hat{k}$.Thus,Statement-$2$ is true.To find a point on the line,set $z = 0$: $3x - 6y = 15 \Rightarrow x - 2y = 5$ and $2x + y = 5$. Solving these,$x = 3, y = -1$. So,$(3, -1, 0)$ is a point on the line.The parametric equations are $x = 3 + 14t, y = -1 + 2t, z = 15t$.Comparing this with Statement-$1$,the $y$-coordinate is given as $1 + 2t$,which is incorrect. Thus,Statement-$1$ is false.

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