If the ends of the hypotenuse of a right-angled triangle are $(0, 1, 2), (1, 2, 0)$ and the locus of the third vertex of the triangle is $x^{2} + y^{2} + z^{2} + a x + b y + c z + d = 0$ , then $a + b + c + d =$

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Solution

The correct option is C $- 4$
Let the third vertex of triangle is $(h, k, l)$
Perpendicular and base in vector form : $(h, k - 1, l - 2)$ and $(h - 1, k - 2, l)$ .
Dot product of two perpendicular lines is zero.
$\Rightarrow (h, k - 1, l - 2) . (h - 1, k - 2, l) = 0$
$\Rightarrow h^{2} - h + k^{2} - 3 k + 2 + l^{2} - 2 l = 0$
$\Rightarrow h^{2} + k^{2} + l^{2} - h - 3 k - 2 l + 2 = 0$
Put $(x, y, z)$ for $(h, k, l)$ it will give locus of third vertex.
So, $a + b + c + d = - 4$

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If the ends of the hypotenuse of a right-angled triangle are (0,1,2),(1,2,0) and the locus of the third vertex of the triangle is x2+y2+z2+ax+by+cz+d=0, then a+b+c+d=

If the ends of the hypotenuse of a right-angled triangle are $(0, 1, 2), (1, 2, 0)$ and the locus of the third vertex of the triangle is $x^{2} + y^{2} + z^{2} + a x + b y + c z + d = 0$ , then $a + b + c + d =$