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(B) The family of planes passing through the intersection of the planes $P_1: x + 2y + 3z - 5 = 0$ and $P_2: 3x + 2y + z - 5 = 0$ is given by $(x + 2y

Vedclass · Accepted Answer

$x - z = 0$

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(B) The family of planes passing through the intersection of the planes $P_1: x + 2y + 3z - 5 = 0$ and $P_2: 3x + 2y + z - 5 = 0$ is given by $(x + 2y + 3z - 5) + \lambda(3x + 2y + z - 5) = 0$. Rearranging the terms,we get $(1 + 3\lambda)x + (2 + 2\lambda)y + (3 + \lambda)z - (5 + 5\lambda) = 0$. The direction ratios of the normal to this plane are $(1 + 3\lambda, 2 + 2\lambda, 3 + \lambda)$. The given line is $\frac{x - 1}{1} = \frac{y - 2}{-1} = \frac{z - 3}{1}$,which has direction ratios $(1, -1, 1)$. Since the plane is parallel to the line,the normal to the plane is perpendicular to the line,so the dot product of their direction ratios is zero: $1(1 + 3\lambda) - 1(2 + 2\lambda) + 1(3 + \lambda) = 0$. $1 + 3\lambda - 2 - 2\lambda + 3 + \lambda = 0$. $2\lambda + 2 = 0 \Rightarrow \lambda = -1$. Substituting $\lambda = -1$ into the equation of the plane: $(x + 2y + 3z - 5) - 1(3x + 2y + z - 5) = 0$. $x + 2y + 3z - 5 - 3x - 2y - z + 5 = 0$. $-2x + 2z = 0 \Rightarrow x - z = 0$.

Equation of the plane containing the line $x + 2y + 3z - 5 = 0 = 3x + 2y + z - 5$ which is parallel to the line $x - 1 = 2 - y = z - 3$ is:

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