Let the area of the triangle with vertices A $\left(1,\alpha \right),B\left(\alpha ,0\right)C(0,\alpha )$ be 4 sq. units. If the points $\left(\alpha ,-\alpha \right),\left(-\alpha ,\alpha \right)and\left({\alpha }^2,\beta \right)$ are collinear, then $\beta$ is equal to

Direction ratio of line (1, -1, -6)
Equation of line (x−3)/1 = (y+4)/-1 = (z+5)/-6 =k
x=k+3, y=−k−4, z=−6k−5
Solving with plane k=−2
⇒x=1, y=−2, z=7
⇒Distance=√ (3−1)²+6²+3²=√49=7

Any point on line (1)
x=α+k
y=1+2k
z=1+3k
Any point on line (2)
x=4+Kβ
y=6+3K
Z=7+3K?
⇒1+2k=6+3K, as the intersect
∴1+3k=7+3K?
⇒K=1, K? =−1
x=α+1; x=4−β
⇒y=3; y=3
z=4; z=4
Equation of plane
x+2y−z=8
⇒α+1+6−4=8 . (i)
and 4−β+6−4=8 . (ii)
Adding (i) and (ii)
α+5−β+12−8=16
α−β+17=24
⇒α−β=7

f (x)= {sinx, 0≤x<π/2; 1, π/2≤x≤π 2+cosx, x>π}
f' (x)= {cosx, 0π}
f' (π/2? ) = 0
f' (π/2? ) = 0
f' (π? ) = 0
f' (π? ) = 0
⇒ f (x) is differentiable in (0, ∞)

Let direction ratio of the normal to the required plane are l, m, n

$\begin{array}{l}\therefore 3l+m-2n=0\\ \therefore 2l-5m-n=0\\ \therefore \frac{l}{-11}=\frac{m}{-1}=\frac{n}{-17}\end{array}$             

Equation of required plane

11 (x – 1) + 1 (y – 2) + 17 (z + 3) = 0

Any point on line  $\frac{x-3}{1}=\frac{y-4}{2}=\frac{z-5}{2}=r$ is P (r + 3, 2r + 4, 2r + 5) lies on x + y + z = 17,

5r + 12 = 17

r = 1

$P\left(4,6,7\right)A\left(1,1,9\right)distAP=\sqrt[]{9+25+4}=\sqrt[]{38}$