The plane containing the line frac{x - 1}{1} | vedclass

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(B) The equation of a plane passing through the point $(x_1, y_1, z_1)$ is $a(x - x_1) + b(y - y_1) + c(z - z_1) = 0$. Since the plane contains the li

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$(1, 0, 5)$

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(B) The equation of a plane passing through the point $(x_1, y_1, z_1)$ is $a(x - x_1) + b(y - y_1) + c(z - z_1) = 0$. Since the plane contains the line $\frac{x - 1}{1} = \frac{y - 2}{2} = \frac{z - 3}{3}$,it passes through $(1, 2, 3)$ and its normal vector $\vec{n} = (a, b, c)$ is perpendicular to the line's direction vector $\vec{v_1} = (1, 2, 3)$. Thus,$a(1) + b(2) + c(3) = 0 \implies a + 2b + 3c = 0$ $(i)$. Since the plane is parallel to the line $\frac{x}{1} = \frac{y}{1} = \frac{z}{4}$,its normal vector is also perpendicular to the line's direction vector $\vec{v_2} = (1, 1, 4)$. Thus,$a(1) + b(1) + c(4) = 0 \implies a + b + 4c = 0$ $(ii)$. Solving $(i)$ and $(ii)$ using cross product: $\frac{a}{8-3} = \frac{b}{3-4} = \frac{c}{1-2} = k$. So,$a = 5k, b = -k, c = -k$. Substituting into the plane equation: $5(x - 1) - 1(y - 2) - 1(z - 3) = 0$. $5x - 5 - y + 2 - z + 3 = 0 \implies 5x - y - z = 0$. Checking the options,for $(1, 0, 5)$: $5(1) - 0 - 5 = 0$. Thus,the plane passes through $(1, 0, 5)$.

The plane containing the line $\frac{x - 1}{1} = \frac{y - 2}{2} = \frac{z - 3}{3}$ and parallel to the line $\frac{x}{1} = \frac{y}{1} = \frac{z}{4}$ passes through the point

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