The equation of the plane which passes throug | vedclass

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(A) The direction ratios of the line joining the points $(3, 4, -1)$ and $(2, -1, 5)$ are given by $(2 - 3, -1 - 4, 5 - (-1)) = (-1, -5, 6)$.Since the

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$x + 5y - 6z + 19 = 0$

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(A) The direction ratios of the line joining the points $(3, 4, -1)$ and $(2, -1, 5)$ are given by $(2 - 3, -1 - 4, 5 - (-1)) = (-1, -5, 6)$.Since the plane is normal to this line,the normal vector to the plane is $\vec{n} = -\hat{i} - 5\hat{j} + 6\hat{k}$.We can also take the normal vector as $\vec{n} = \hat{i} + 5\hat{j} - 6\hat{k}$.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 point $(2, -3, 1)$ and the normal vector $(1, 5, -6)$,we get:$1(x - 2) + 5(y - (-3)) - 6(z - 1) = 0$$x - 2 + 5y + 15 - 6z + 6 = 0$$x + 5y - 6z + 19 = 0$.

The equation of the plane which passes through $(2, -3, 1)$ and is normal to the line joining the points $(3, 4, -1)$ and $(2, -1, 5)$ is given by

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