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(A) The equation of a plane passing through $(2,-3,4)$ with normal vector $\vec{n} = (a, b, c)$ is given by $a(x-2) + b(y+3) + c(z-4) = 0$, which simp

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$\frac{5}{11}$

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(A) The equation of a plane passing through $(2,-3,4)$ with normal vector $\vec{n} = (a, b, c)$ is given by $a(x-2) + b(y+3) + c(z-4) = 0$, which simplifies to $ax + by + cz - 2a + 3b - 4c = 0$ ... $(i)$Since the plane is perpendicular to both $2x-3y+5z=2$ and $x+y+2z=3$, its normal vector $\vec{n}$ must be parallel to the cross product of the normals of the given planes, $\vec{n_1} = (2, -3, 5)$ and $\vec{n_2} = (1, 1, 2)$.$\vec{n} = \vec{n_1} 	imes \vec{n_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 2 & -3 & 5 \ 1 & 1 & 2 \end{vmatrix} = \hat{i}(-6-5) - \hat{j}(4-5) + \hat{k}(2+3) = -11\hat{i} + 1\hat{j} + 5\hat{k}$.Thus, $(a, b, c) = (-11, 1, 5)$.Substituting these into Eq. $(i)$: $-11x + y + 5z - 2(-11) + 3(1) - 4(5) = 0$$-11x + y + 5z + 22 + 3 - 20 = 0$$-11x + y + 5z + 5 = 0 \Rightarrow 11x - y - 5z = 5$Dividing by $11$, we get $x - \frac{1}{11}y - \frac{5}{11}z = \frac{5}{11}$.Comparing this with $x + py + qz = r$, we find $r = \frac{5}{11}$.

If the equation of the plane passing through the point $(2,-3,4)$ and perpendicular to both the planes $2x-3y+5z=2$ and $x+y+2z=3$ is $x+py+qz=r$, then $r$ is equal to

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