The equation of a plane passing through (-1, | vedclass

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(C) Let the normal vector to the plane be $\vec{n} = (a, b, c)$. Since the normal makes equal angles with the coordinate axes,the direction cosines ar

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$x+y+z-4=0$

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(C) Let the normal vector to the plane be $\vec{n} = (a, b, c)$. Since the normal makes equal angles with the coordinate axes,the direction cosines are equal,i.e.,$\cos \alpha = \cos \beta = \cos \gamma$. This implies $a = b = c$. Thus,the equation of the plane is of the form $x + y + z = d$. Since the plane passes through the point $(-1, 2, 3)$,we substitute these coordinates into the equation: $-1 + 2 + 3 = d \Rightarrow d = 4$. Substituting the value of $d$ back into the equation,we get: $x + y + z = 4$ $x + y + z - 4 = 0$.

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

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