If the lines L_1 : frac{x - 1}{-3} = frac{y - | vedclass

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(B) Step $1$: Find $k$ using the condition of perpendicularity. The direction vectors are $\vec{v_1} = (-3, 2k, 2)$ and $\vec{v_2} = (3k, 1, -5)$. Sin

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$602x - 1155y - 747z + 3949 = 0$

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(B) Step $1$: Find $k$ using the condition of perpendicularity. The direction vectors are $\vec{v_1} = (-3, 2k, 2)$ and $\vec{v_2} = (3k, 1, -5)$. Since $L_1 \perp L_2$, $\vec{v_1} \cdot \vec{v_2} = 0$.Step $2$: $(-3)(3k) + (2k)(1) + (2)(-5) = 0 \implies -9k + 2k - 10 = 0 \implies -7k = 10 \implies k = -\frac{10}{7}$.Step $3$: The normal vector $\vec{n}$ to the plane is $\vec{v_1} 	imes \vec{v_2}$. Since $L_1 \perp L_2$, $\vec{n} = \vec{v_1} 	imes \vec{v_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ -3 & 2k & 2 \ 3k & 1 & -5 \end{vmatrix} = \hat{i}(-10k - 2) - \hat{j}(15 - 6k) + \hat{k}(-3 - 6k^2)$.Step $4$: Substitute $k = -\frac{10}{7}$: $\vec{n} = \hat{i}(-10(-\frac{10}{7}) - 2) - \hat{j}(15 - 6(-\frac{10}{7})) + \hat{k}(-3 - 6(-\frac{10}{7})^2) = \hat{i}(\frac{100-14}{7}) - \hat{j}(\frac{105+60}{7}) + \hat{k}(\frac{-147-600}{49}) = \frac{86}{7}\hat{i} - \frac{165}{7}\hat{j} - \frac{747}{49}\hat{k}$.Step $5$: Multiply by $49$ to simplify: $\vec{n} = 602\hat{i} - 1155\hat{j} - 747\hat{k}$.Step $6$: The plane passes through $(1, 2, 3)$. Equation: $602(x-1) - 1155(y-2) - 747(z-3) = 0 \implies 602x - 1155y - 747z - 602 + 2310 + 2241 = 0 \implies 602x - 1155y - 747z + 3949 = 0$.

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