A plane pi passing through the point (1,1,1) | vedclass

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(A) The direction ratios of the line joining the points $(6,3,2)$ and $(1,-4,-9)$ are $(6-1, 3-(-4), 2-(-9)) = (5, 7, 11)$.Since the plane is perpendi

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(A) The direction ratios of the line joining the points $(6,3,2)$ and $(1,-4,-9)$ are $(6-1, 3-(-4), 2-(-9)) = (5, 7, 11)$.Since the plane is perpendicular to this line, the direction ratios of the normal to the plane are $(a, b, c) = (5, 7, 11)$.The equation of the plane passing through $(x_0, y_0, z_0) = (1, 1, 1)$ with normal vector $(a, b, c) = (5, 7, 11)$ is given by $a(x-x_0) + b(y-y_0) + c(z-z_0) = 0$.Substituting the values, we get $5(x-1) + 7(y-1) + 11(z-1) = 0$.Expanding this, we get $5x - 5 + 7y - 7 + 11z - 11 = 0$, which simplifies to $5x + 7y + 11z - 23 = 0$.Comparing this with $ax + by + cz - 23 = 0$, we find $a=5, b=7, c=11$.Therefore, $a+b-c = 5+7-11 = 1$.

$A$ plane $\pi$ passing through the point $(1,1,1)$ is perpendicular to the line joining the points $(6,3,2)$ and $(1,-4,-9)$. If $ax+by+cz-23=0$ is the equation of the plane $\pi$, then $a+b-c=$

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