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(D) The two given planes are mutually perpendicular,so the dot product of their normals is zero:$2(3k) + k(-k) + (-5)(1) = 0$$6k - k^2 - 5 = 0 \Righta

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$7$

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(D) The two given planes are mutually perpendicular,so the dot product of their normals is zero:$2(3k) + k(-k) + (-5)(1) = 0$$6k - k^2 - 5 = 0 \Rightarrow k^2 - 6k + 5 = 0$$(k - 1)(k - 5) = 0 \Rightarrow k = 1$ or $k = 5$.Given that $k The equation of the plane passing through the intersection of the two planes is:$(2x + y - 5z - 1) + \lambda(3x - y + z - 5) = 0$$(2 + 3\lambda)x + (1 - \lambda)y + (-5 + \lambda)z = 1 + 5\lambda$.The intercept on the $x$-axis is $1$,so setting $y = 0$ and $z = 0$:$\frac{1 + 5\lambda}{2 + 3\lambda} = 1 \Rightarrow 1 + 5\lambda = 2 + 3\lambda \Rightarrow 2\lambda = 1 \Rightarrow \lambda = \frac{1}{2}$.To find the intercept on the $y$-axis,set $x = 0$ and $z = 0$:$(1 - \lambda)y = 1 + 5\lambda \Rightarrow y = \frac{1 + 5(1/2)}{1 - 1/2} = \frac{1 + 2.5}{0.5} = \frac{3.5}{0.5} = 7$.

If the plane $P$ passes through the intersection of two mutually perpendicular planes $2x + ky - 5z = 1$ and $3kx - ky + z = 5$,where $k < 3$,and intercepts a unit length on the positive $x$-axis,then the intercept made by the plane $P$ on the $y$-axis is

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