Let \(\vec a = \hat i + \hat j + \hat k\)
\(\vec b = 2\hat i + 4\hat j - 5\hat k\)
\(\vec c = \lambda\hat i + 2\hat j + 3 \hat k\)
\((\vec b + \vec c) = (2 + \lambda)\hat i + (4 + 2)\hat j + (-5 +3)\hat k\)
\(= (2 + \lambda)\hat i + 6\hat j - 2\hat k\)
Let \(\vec r\) be unit vector along \((\vec b + \vec c)\)
\(\hat r = \frac 1{|\vec b + \vec c|} \times (\vec b \times \vec c)\)
\(\hat r = \frac 1{\sqrt{(2 + \lambda)^2 + 36 + 4}}\times [(2 + \lambda)\hat i + 6\hat j - 2\hat k)]\)
\(\hat r = \frac 1{\sqrt{ \lambda^2 +4\lambda+ 44 }}\times [(2 + \lambda)\hat i + 6\hat j - 2\hat k]\)
Now \(\vec a.\vec r = 1\) (given)
\((\hat i + \hat j + \hat k).\left(\frac 1{\sqrt{\lambda^2 + 4\lambda + 44}}\times (2 + \lambda)\hat i + 6\hat j - 2\hat k\right) = 1\)
\((\hat i + \hat j + \hat k). ((2 + \lambda)\hat y+ 6\hat j - 2\hat k)= \sqrt{\lambda^2 + 4\lambda + 44}\)
\((\lambda + 2) + 6 + 1.(-2) = \sqrt{\lambda^2 + 4\lambda + 44}\)
\(\lambda + 2 + 6 -2= \sqrt{\lambda^2 + 4\lambda + 44}\)
\((\lambda + 6)^2 = \lambda^2 + 4\lambda + 44\)
\(\lambda^2 + 12 \lambda + 36 = \lambda^2 + 4\lambda + 44\)
\(8\lambda = 8\)
\(\lambda = 1\)
Hence, the value of λ is 1.
