Let the plane passing through the point (-1, 0, -2) and perpendicular to each of the planes $2x+y-z=2$ and $x-y-z=3$ be $ax+by+cz+8=0$. Then the value of $a+b+c$ is equal to :
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The normal vector of a plane that is perpendicular to two other planes is parallel to the cross product of the normal vectors of those two planes. This is a standard method for finding the orientation of such a plane.
Let the equation of the required plane be $P$. The normal vector to the plane $P_1: 2x+y-z=2$ is $\vec{n}_1 = 2\hat{i} + \hat{j} - \hat{k}$. The normal vector to the plane $P_2: x-y-z=3$ is $\vec{n}_2 = \hat{i} - \hat{j} - \hat{k}$. The required plane $P$ is perpendicular to both $P_1$ and $P_2$. This means the normal vector of $P$, let's call it $\vec{n}$, must be perpendicular to both $\vec{n}_1$ and $\vec{n}_2$. Therefore, $\vec{n}$ is parallel to the cross product $\vec{n}_1 \times \vec{n}_2$. $= \hat{i}((-1)( -1) - (-1)(1)) - \hat{j}((-1)(2) - (-1)(1)) + \hat{k}((2)(-1) - (1)(1))$ $= \hat{i}(-1-1) - \hat{j}(-2+1) + \hat{k}(-2-1) = -2\hat{i} + \hat{j} - 3\hat{k}$. So the direction ratios of the normal to the required plane are (-2, 1, -3). The equation of the plane is of the form $-2x + 1y - 3z + d = 0$. The plane passes through the point (-1, 0, -2). We substitute these coordinates to find d. $-2(-1) + 1(0) - 3(-2) + d = 0$ $2 + 0 + 6 + d = 0 \implies d = -8$. The equation of the plane is $-2x + y - 3z - 8 = 0$. We are given the equation in the form $ax+by+cz+8=0$. To match the constant term, we multiply our equation by -1: $2x - y + 3z + 8 = 0$. Comparing this with $ax+by+cz+8=0$, we get: $a=2, b=-1, c=3$. The value of $a+b+c$ is $2 + (-1) + 3 = 4$.
