The equation of the plane perpendicular to the line x-2 1 - y-1 2 - z-1 -1 and passing through the point 3-0-5 is Ax-By-Cz-2-0- then A-B-C is-

If equation of the plane through the straight line $\frac{x - 1}{2} = \frac{y + 2}{- 3} = \frac{z}{5}$ and perpendicular to the plane $x - y + z + 2 = 0$ is $a x - b y + c z + 4 = 0,$ then find the value of $a^{2} + b^{2} + c$

Q. Let the plane passing through the point

(- 1, 0, - 2)

and perpendicular to each of the planes

2 x + y - z = 2

and

x - y - z = 3

a x + b y + c z + 8 = 0

. Then the value of

a + b + c

is equal to

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The equation of the plane perpendicular to the line x+21=y−12=z+1−1 and passing through the point (3,0,5) is Ax+By+Cz+2=0, then (A+B+C) is-

The correct option is A 2Given line Ax+By+Cz+2=0 is passing through point (3,0,5)Hence 3A+5C+2=0---(1)and The given line is also perpendicular to line x+21=y−12=z+1−1Hence a1a2+b1b2+c1c2=0A+2B−C=0Let B=0A=C3A+5A=−2A=−14=CA+B+C=−14−14A+B+C=−12

The equation of the plane perpendicular to the line $\frac{x + 2}{1} = \frac{y - 1}{2} = \frac{z + 1}{- 1}$ and passing through the point $(3, 0, 5)$ is $A x + B y + C z + 2 = 0$ , then $(A + B + C)$ is-