The value of k for which the planes 3x - 6y - | vedclass

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(A) Two planes $a_1x + b_1y + c_1z = d_1$ and $a_2x + b_2y + c_2z = d_2$ are perpendicular if their normal vectors $\vec{n_1} = (a_1, b_1, c_1)$ and $

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(A) Two planes $a_1x + b_1y + c_1z = d_1$ and $a_2x + b_2y + c_2z = d_2$ are perpendicular if their normal vectors $\vec{n_1} = (a_1, b_1, c_1)$ and $\vec{n_2} = (a_2, b_2, c_2)$ are perpendicular,i.e.,$\vec{n_1} \cdot \vec{n_2} = 0$.Here,the normal vectors are $\vec{n_1} = (3, -6, -2)$ and $\vec{n_2} = (2, 1, -k)$.For the planes to be perpendicular,the dot product of their normal vectors must be zero:$(3)(2) + (-6)(1) + (-2)(-k) = 0$$6 - 6 + 2k = 0$$0 + 2k = 0$$2k = 0$$k = 0$.

The value of $k$ for which the planes $3x - 6y - 2z = 7$ and $2x + y - kz = 5$ are perpendicular to each other,is

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