A plane passing through (-1, 2, 3) and whose | vedclass

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(C) The equation of a plane passing through the point $(x_0, y_0, z_0)$ with normal vector $\vec{n} = \langle a, b, c \rangle$ is given by $a(x-x_0) +

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$x+y+z-4=0$

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(C) The equation of a plane passing through the point $(x_0, y_0, z_0)$ with normal vector $\vec{n} = \langle a, b, c \rangle$ is given by $a(x-x_0) + b(y-y_0) + c(z-z_0) = 0$.Given the point is $(-1, 2, 3)$,the equation is $a(x+1) + b(y-2) + c(z-3) = 0$.The normal makes equal angles $\alpha$ with the coordinate axes,so the direction cosines are $\cos \alpha, \cos \alpha, \cos \alpha$.Since $\cos^2 \alpha + \cos^2 \alpha + \cos^2 \alpha = 1$,we have $3 \cos^2 \alpha = 1$,which implies $\cos \alpha = \pm \frac{1}{\sqrt{3}}$.Thus,the direction ratios $\langle a, b, c \rangle$ can be taken as $\langle 1, 1, 1 \rangle$.Substituting these into the plane equation: $1(x+1) + 1(y-2) + 1(z-3) = 0$.Simplifying,we get $x + 1 + y - 2 + z - 3 = 0$,which results in $x + y + z - 4 = 0$.

$A$ plane passing through $(-1, 2, 3)$ and whose normal makes equal angles with the coordinate axes is

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