The line drawn from (4, -1, 2) to the point ( | vedclass

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(A) The line passes through the points $P(4, -1, 2)$ and $Q(-3, 2, 3)$.The direction ratios of this line are given by $(x_2 - x_1, y_2 - y_1, z_2 - z_

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$7x - 3y - z + 89 = 0$

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(A) The line passes through the points $P(4, -1, 2)$ and $Q(-3, 2, 3)$.The direction ratios of this line are given by $(x_2 - x_1, y_2 - y_1, z_2 - z_1) = (-3 - 4, 2 - (-1), 3 - 2) = (-7, 3, -1)$.Since the line is perpendicular to the plane,the direction ratios of the normal to the plane are the same as the direction ratios of the line,which are $(-7, 3, -1)$.Alternatively,we can use the normal vector $\vec{n} = (7, -3, 1)$ by multiplying by $-1$.The plane passes through the point $R(-10, 5, 4)$.The equation of a plane passing through $(x_0, y_0, z_0)$ with normal vector $(a, b, c)$ is $a(x - x_0) + b(y - y_0) + c(z - z_0) = 0$.Substituting the values,we get $7(x - (-10)) - 3(y - 5) - 1(z - 4) = 0$.$7(x + 10) - 3(y - 5) - (z - 4) = 0$.$7x + 70 - 3y + 15 - z + 4 = 0$.$7x - 3y - z + 89 = 0$.

The line drawn from $(4, -1, 2)$ to the point $(-3, 2, 3)$ meets a plane at right angles at the point $(-10, 5, 4)$. Find the equation of the plane.

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