A plane containing two lines whose direction | vedclass

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(A) The normal vector $\vec{n}$ to the plane is the cross product of the direction vectors of the two lines, $\vec{v_1} = (-1, 2, 1)$ and $\vec{v_2} =

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$5$

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(A) The normal vector $\vec{n}$ to the plane is the cross product of the direction vectors of the two lines, $\vec{v_1} = (-1, 2, 1)$ and $\vec{v_2} = (1, 3, 2)$.$\vec{n} = \vec{v_1} 	imes \vec{v_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ -1 & 2 & 1 \ 1 & 3 & 2 \end{vmatrix} = \hat{i}(4-3) - \hat{j}(-2-1) + \hat{k}(-3-2) = \hat{i} + 3\hat{j} - 5\hat{k}$.Since the plane passes through the point $(2, 1, k)$, its equation is given by $1(x-2) + 3(y-1) - 5(z-k) = 0$.Substituting the point $(3, -1, 4)$ into the equation:$1(3-2) + 3(-1-1) - 5(4-k) = 0$.$1(1) + 3(-2) - 20 + 5k = 0$.$1 - 6 - 20 + 5k = 0$.$-25 + 5k = 0$.$5k = 25 \Rightarrow k = 5$.

$A$ plane containing two lines whose direction ratios are $(-1, 2, 1)$ and $(1, 3, 2)$ passes through the point $(2, 1, k)$. If this plane also passes through the point $(3, -1, 4)$, then $k=$

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