The distance of the plane 3x + 4y + 5z + 19 = | vedclass

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(C) The equation of the line passing through the point $(1, -1, 1)$ and parallel to the line with direction ratios $(2, 3, 1)$ is given by:$\frac{x -

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$\sqrt{14}$

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(C) The equation of the line passing through the point $(1, -1, 1)$ and parallel to the line with direction ratios $(2, 3, 1)$ is given by:$\frac{x - 1}{2} = \frac{y + 1}{3} = \frac{z - 1}{1} = r$Any point on this line can be represented as $(2r + 1, 3r - 1, r + 1)$.Since this point lies on the plane $3x + 4y + 5z + 19 = 0$,it must satisfy the equation of the plane:$3(2r + 1) + 4(3r - 1) + 5(r + 1) + 19 = 0$$6r + 3 + 12r - 4 + 5r + 5 + 19 = 0$$23r + 23 = 0$$r = -1$The point of intersection is $(2(-1) + 1, 3(-1) - 1, -1 + 1) = (-1, -4, 0)$.The distance between the points $(1, -1, 1)$ and $(-1, -4, 0)$ is:$d = \sqrt{(-1 - 1)^2 + (-4 - (-1))^2 + (0 - 1)^2}$$d = \sqrt{(-2)^2 + (-3)^2 + (-1)^2}$$d = \sqrt{4 + 9 + 1} = \sqrt{14}$

The distance of the plane $3x + 4y + 5z + 19 = 0$ from the point $(1, -1, 1)$ measured along a line parallel to the line with direction ratios $2, 3, 1$ is

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