The equation of the plane passing through (1, | vedclass

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(B) The equation of any plane passing through $(1, 1, 1)$ is given by $a(x - 1) + b(y - 1) + c(z - 1) = 0 \dots (i)$.Since the plane passes through $(

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$x + y - z - 1 = 0$

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(B) The equation of any plane passing through $(1, 1, 1)$ is given by $a(x - 1) + b(y - 1) + c(z - 1) = 0 \dots (i)$.Since the plane passes through $(1, -1, -1)$,we substitute these coordinates into $(i)$:$a(1 - 1) + b(-1 - 1) + c(-1 - 1) = 0 \implies 0a - 2b - 2c = 0 \implies b + c = 0 \dots (ii)$.The plane is perpendicular to $2x - y + z + 5 = 0$,so the normal vectors are perpendicular. The normal vector of the required plane is $\vec{n} = (a, b, c)$ and the normal of the given plane is $\vec{n_1} = (2, -1, 1)$.Thus,$2a - b + c = 0 \dots (iii)$.From $(ii)$,$c = -b$. Substituting this into $(iii)$:$2a - b - b = 0 \implies 2a - 2b = 0 \implies a = b$.Let $a = 1$,then $b = 1$ and $c = -1$.Substituting these values into $(i)$:$1(x - 1) + 1(y - 1) - 1(z - 1) = 0 \implies x - 1 + y - 1 - z + 1 = 0 \implies x + y - z - 1 = 0$.

The equation of the plane passing through $(1, 1, 1)$ and $(1, -1, -1)$ and perpendicular to $2x - y + z + 5 = 0$ is

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