The point of intersection of the line frac{x | vedclass

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(B) Let the given line be $\frac{x - 1}{3} = \frac{y + 2}{4} = \frac{z - 3}{-2} = k$. Any point on the line is given by $(x, y, z) = (3k + 1, 4k - 2,

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$(10, 10, -3)$

Vedclass · Answer

(B) Let the given line be $\frac{x - 1}{3} = \frac{y + 2}{4} = \frac{z - 3}{-2} = k$. Any point on the line is given by $(x, y, z) = (3k + 1, 4k - 2, -2k + 3)$. Since this point lies on the plane $2x - y + 3z - 1 = 0$,it must satisfy the plane equation: $2(3k + 1) - (4k - 2) + 3(-2k + 3) - 1 = 0$. Expanding the equation: $6k + 2 - 4k + 2 - 6k + 9 - 1 = 0$. Simplifying: $-4k + 12 = 0$,which gives $4k = 12$,so $k = 3$. Substituting $k = 3$ back into the point coordinates: $x = 3(3) + 1 = 10$,$y = 4(3) - 2 = 10$,$z = -2(3) + 3 = -3$. Thus,the point of intersection is $(10, 10, -3)$.

The point of intersection of the line $\frac{x - 1}{3} = \frac{y + 2}{4} = \frac{z - 3}{-2}$ and the plane $2x - y + 3z - 1 = 0$ is

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