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(B) The equation of a plane passing through $(1, 1, 1)$ is $a(x-1) + b(y-1) + c(z-1) = 0$.Since the plane is parallel to the lines with direction rati

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$\frac{9}{2}$

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(B) The equation of a plane passing through $(1, 1, 1)$ is $a(x-1) + b(y-1) + c(z-1) = 0$.Since the plane is parallel to the lines with direction ratios $(1, 0, -1)$ and $(-1, 1, 0)$,the normal vector $(a, b, c)$ must be perpendicular to both direction vectors.Thus,$a(1) + b(0) + c(-1) = 0 \Rightarrow a = c$ and $a(-1) + b(1) + c(0) = 0 \Rightarrow a = b$.Therefore,$a = b = c$. Setting $a = b = c = 1$,the equation of the plane becomes $1(x-1) + 1(y-1) + 1(z-1) = 0$,which simplifies to $x + y + z = 3$.Dividing by $3$,we get the intercept form $\frac{x}{3} + \frac{y}{3} + \frac{z}{3} = 1$.The intercepts are $A(3, 0, 0)$,$B(0, 3, 0)$,and $C(0, 0, 3)$.The volume of the tetrahedron $OABC$ is given by $V = \frac{1}{6} |x_A y_B z_C| = \frac{1}{6} |3 	imes 3 	imes 3| = \frac{27}{6} = \frac{9}{2} 	ext{ cubic units}$.

$A$ plane is parallel to two lines whose direction ratios are $1, 0, -1$ and $-1, 1, 0$ and it contains the point $(1, 1, 1)$. If it cuts the coordinate axes at $A, B, C$,then the volume of the tetrahedron $OABC$ (in cubic units) is

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