If the line -x+1 3 = -y-2 -2k = z+3 2 and -1+x 3k = -1 +y 1 = -z+6 5 are perpendicular, then the value of k is:

Standard symmetric form of a line:
$$\frac{x - x_1}{a} = \frac{y - y_1}{b} = \frac{z - z_1}{c}$$ax−x1​​=by−y1​​=cz−z1​​

The given equations are:
$$\frac{-x + 1}{3} = \frac{-y - 2}{-2k} = \frac{z + 3}{2}$$3−x+1​=−2k−y−2​=2z+3​
and
$$\frac{-1 + x}{3k} = \frac{-1 + y}{1} = \frac{-z + 6}{5}$$3k−1+x​=1−1+y​=5−z+6​

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$$\frac{x - 1}{-3} = \frac{y + 2}{2k} = \frac{z + 3}{2}$$−3x−1​=2ky+2​=2z+3​

The direction vector for the first line is:
$$\vec{d_1} = \langle -3, 2k, 2 \rangle$$d1​​=⟨−3,2k,2⟩

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$$\frac{x - 1}{3k} = \frac{y - 1}{1} = \frac{z - 6}{-5}$$3kx−1​=1y−1​=−5z−6​

The direction vector for the second line is:
$$\vec{d_2} = \langle 3k, 1, -5 \rangle$$d2​​=⟨3k,1,−5⟩

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$$\vec{d_1} \cdot \vec{d_2} = 0$$d1​​⋅d2​​=0

$$(-3)(3k) + (2k)(1) + (2)(-5) = 0$$(−3)(3k)+(2k)(1)+(2)(−5)=0

$$-9k + 2k - 10 = 0$$−9k+2k−10=0

$$-7k - 10 = 0$$−7k−10=0

$$-7k = 10$$−7k=10

$$k = -\frac{10}{7}$$k=−710​