Let a = i + 2j - 3k and b = 2i - 3j + 5k. If r x a = b x r, r.(αi+2j+k) = 3 and r.(2i + 5j - αk) = -1, α ∈ R, then the value of α + |r|² is equal to :

<p>Given r x a = b x r, which means r x a + r x b = 0, so r x (a+b) = 0.<br>This implies r is parallel to (a+b). So, r = λ (a+b).<br>a = I + 2j - 3k, b = 2i - 3j + 5k<br>a+b = 3i - j + 2k.<br>r = λ (3i - j + 2k).<br>Given r ⋅ (αi + 2j + k) = 3. The OCR is unclear, but the equation appears to be r ⋅ (αi + 2j + k) = 3. The solution works with r ⋅ (αi + 2j + k) = 3, but theOCR says r. (ai+2j+k). Let's assume it's α.<br>λ (3i - j + 2k) ⋅ (αi + 2j + k) = 3<br>λ (3α - 2 + 2) = 3 =&gt; λα = 1.<br>Given r ⋅ (2i + 5j - αk) = -1.<br>λ (3i - j + 2k) ⋅ (2i + 5j - αk) = -1<br>λ (6 - 5 - 2α) = -1 =&gt; λ (1 - 2α) = -1.<br>From λα = 1, α = 1/λ.<br>λ (1 - 2/λ) = -1 =&gt; λ - 2 = -1 =&gt; λ = 1.<br>If λ = 1, then α = 1.<br>r = 1 * (3i - j + 2k) = 3i - j + 2k.<br>|r| = √ (3² + (-1)² + 2²) = √ (9+1+4) = √14.<br>We need to find α + |r|².<br>α + |r|² = 1 + (√14)² = 1 + 14 = 15.</p>

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